STRAIGHT LINES MCQs for JEE MATHEMATICS — Mixed difficulty, Page 6

151 Easy

151 of 530

Application +4 | -1

The set of all possible values of

\theta

in the interval

\left( {0,\,\,\pi } \right)\,

for which the points

\left( {1,\,\,2} \right)

and

\left( {\sin \theta ,\,\,\,\cos \theta } \right)\,

lie on the same side of the line

x + y = 1

is

View answer and explanation

Correct option: A

\left( {0,\,\,\,\frac{\pi }{2}} \right)\,\,

Option A (

\left( {0,\\,\\,\frac{\pi }{2}} \right)\,\,

) is correct because it represents the interval where both points lie on the same side of the line x + y = 1, derived from the inequality sin(theta) + cos(theta) > 1.

Solution approach

Correct Answer + Core Reason: $\left( {0,\,\,\,\frac{\pi }{2}} \right)\,\,$ . The interval is found by ensuring that the points (1,2) and (sin θ, cos θ) lie on the same side of the line x + y = 1.

Step-by-Step Solution:

Step 1: The points (1,2) and (sin θ, cos θ) lie on the same side of the line x + y = 1. Therefore, (1 + 2 - 1)(sin θ + cos θ - 1) > 0.

Step 2: Simplify: (2)(sin θ + cos θ - 1) > 0, which means sin θ + cos θ > 1.

Step 3: Rewrite sin θ + cos θ as $\sqrt 2 \sin \left( {\theta + \frac{\pi }{4}} \right) > 1$ . Therefore, $\sin \left( {\theta + \frac{\pi }{4}} \right) > \frac{1}{{\sqrt 2 }}$ .

Step 4: Find the general solution for $\theta + \frac{\pi }{4}$ : $2n\pi + \frac{\pi }{4} < \theta + \frac{\pi }{4} < 2n\pi + \frac{{3\pi }}{4}$ , where n is an integer.

Step 5: Simplify: $2n\pi < \theta < 2n\pi + \frac{\pi }{2}$ .

Step 6: Since $\theta \in \left( {0,\,\,\pi } \right)$ , let n = 0. Then, $0 < \theta < \frac{\pi }{2}$ .

Why other options are wrong:
- Option B: This option is incorrect because it does not satisfy the condition sin θ + cos θ > 1 for all values in the interval.
- Option C: This option is incorrect because it includes values where the points lie on opposite sides of the line.
- Option D: This option is incorrect because it does not include all values where the points lie on the same side of the line.

Key Concepts:
- Equation of a line
- Trigonometric inequalities
- General solutions of trigonometric equations

Memory Aids:
- Remember that points on the same side of a line will have the same sign when substituted into the line's equation.

Quick Tips:
- Always check the interval of θ when finding the solution.
- Use trigonometric identities to simplify the inequality.

152 Easy

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Analyzing +4 | -1

The straight lines

x = y,\;\;x - 2y = \;3

and

x + 2y = - 3

form a triangle, which is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the side lengths of the triangle formed by the given lines are not equal, nor do they satisfy the Pythagorean theorem, indicating it is neither isosceles, equilateral, nor right-angled.

Solution approach

Correct Answer + Core Reason: None of these. The triangle formed by the given lines is neither isosceles, equilateral, nor right-angled because the side lengths are different and do not satisfy the Pythagorean theorem.

Step-by-Step Solution:

Step 1: Find the intersection points of the lines: $x = y$ , $x - 2y = 3$ , and $x + 2y = -3$ .

Step 2: Intersection of $x = y$ and $x - 2y = 3$ : Substituting $x = y$ into $x - 2y = 3$ gives $y - 2y = 3$ , so $y = -3$ and $x = -3$ . Point A is (-3, -3).

Step 3: Intersection of $x = y$ and $x + 2y = -3$ : Substituting $x = y$ into $x + 2y = -3$ gives $y + 2y = -3$ , so $3y = -3$ , $y = -1$ and $x = -1$ . Point B is (-1, -1).

Step 4: Intersection of $x - 2y = 3$ and $x + 2y = -3$ : Adding the two equations gives $2x = 0$ , so $x = 0$ . Substituting $x = 0$ into $x - 2y = 3$ gives $-2y = 3$ , so $y = -\frac{3}{2}$ . Point C is (0, - $\frac{3}{2}$ ).

Step 5: Calculate the side lengths: $AB = \sqrt {(-1 - (-3))^2 + (-1 - (-3))^2} = \sqrt {2^2 + 2^2} = \sqrt 8 = 2\sqrt 2$ .
$AC = \sqrt {(0 - (-3))^2 + (-\frac{3}{2} - (-3))^2} = \sqrt {3^2 + (\frac{3}{2})^2} = \sqrt {9 + \frac{9}{4}} = \sqrt {\frac{45}{4}} = \frac{{3\sqrt 5 }}{2}$ .
$BC = \sqrt {(0 - (-1))^2 + (-\frac{3}{2} - (-1))^2} = \sqrt {1^2 + (-\frac{1}{2})^2} = \sqrt {1 + \frac{1}{4}} = \sqrt {\frac{5}{4}} = \frac{{\sqrt 5 }}{2}$ .

Step 6: Check if the triangle is isosceles: $AB \ne AC \ne BC$ , so it is not isosceles.

Step 7: Check if the triangle is equilateral: $AB \ne AC \ne BC$ , so it is not equilateral.

Step 8: Check if the triangle is right-angled: $AB^2 + BC^2 = (2\sqrt 2)^2 + (\frac{{\sqrt 5 }}{2})^2 = 8 + \frac{5}{4} = \frac{37}{4} \ne (\frac{{3\sqrt 5 }}{2})^2 = \frac{45}{4}$ . Therefore, it is not right-angled.

Why other options are wrong:
- Option A (Isosceles): This is incorrect because the side lengths are not equal.
- Option B (Equilateral): This is incorrect because the side lengths are not equal.
- Option C (Right angled): This is incorrect because the side lengths do not satisfy the Pythagorean theorem.

Key Concepts:
- Finding intersection points of lines
- Distance formula
- Types of triangles (isosceles, equilateral, right-angled)
- Pythagorean theorem

Memory Aids:
- Distance formula: $d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}$
- Pythagorean theorem: $a^2 + b^2 = c^2$

Quick Tips:
- Find the intersection points of the lines to determine the vertices of the triangle.
- Calculate the side lengths using the distance formula.
- Check the side lengths to determine the type of triangle.

153 Easy

153 of 530

Application +4 | -1

If the points

A(\lambda, 2 \lambda), B(3 \lambda, 3 \lambda)

and

C(3,1)

are collinear, then

\lambda=

View answer and explanation

Correct option: B

-1 / 3

Option B (-1/3) is correct because it satisfies the condition for the three points to be collinear.

Solution approach

Correct Answer + Core Reason: -1/3. The collinearity condition leads to this value of lambda.

Step-by-Step Solution:

Step 1: Given points $A(\lambda, 2\lambda)$ , $B(3\lambda, 3\lambda)$ , and $C(3,1)$ .

Step 2: For the points to be collinear, the determinant of the matrix formed by their coordinates must be zero.

Step 3: $\begin{vmatrix} \lambda & 2\lambda & 1 \\ 3\lambda & 3\lambda & 1 \\ 3 & 1 & 1 \end{vmatrix} = 0$ .

Step 4: Expanding the determinant, we get $\lambda(3\lambda - 1) - 2\lambda(3\lambda - 3) + 1(3\lambda - 9\lambda) = 0$ .

Step 5: Simplifying, we get $3\lambda^2 - \lambda - 6\lambda^2 + 6\lambda + 3\lambda - 9\lambda = 0$ .

Step 6: Further simplification gives $-3\lambda^2 - \lambda = 0$ .

Step 7: Factoring out $\lambda$ , we get $\lambda(-3\lambda - 1) = 0$ .

Step 8: Thus, $\lambda = 0$ or $-3\lambda - 1 = 0$ , which gives $\lambda = -\frac{1}{3}$ .

Step 9: Since the problem implies $\lambda$ is non-zero, $\lambda = -\frac{1}{3}$ .

Why other options are wrong:
- Option A: This option is incorrect because it does not satisfy the collinearity condition.
- Option C: This option is incorrect because it does not satisfy the collinearity condition.
- Option D: This option is incorrect because it does not satisfy the collinearity condition.

Key Concepts:
- Collinearity of points
- Determinant of a matrix

Memory Aids:
- Remember that for collinear points, the determinant of the matrix formed by their coordinates is zero.

Quick Tips:
- Use the determinant method to determine the condition for collinearity.

154 Medium

154 of 530

Analyzing +4 | -1

ax + by - {a^2} = 0,\;

where

a,b

are non-zero, is the equation to the straight line perpendicular to a line

l

and passing through the point where

l

crosses the x-axis. Then, equation to the line

l

is

View answer and explanation

Correct option: B

\frac{x}{a} - \frac{y}{b} = 1

Option B (

\frac{x}{a} - \frac{y}{b} = 1

) is correct because it represents the equation of the line

l

that is perpendicular to the given line and passes through the x-intercept of the given line.

Solution approach

Correct Answer + Core Reason: $\frac{x}{a} - \frac{y}{b} = 1$ . The equation of the line is found by using the perpendicularity condition and the x-intercept of the given line.

Step-by-Step Solution:

Step 1: Given equation of the line: $ax + by - {a^2} = 0$ .

Step 2: Find the slope of the given line: $ax + by - {a^2} = 0 \Rightarrow by = -ax + a^2 \Rightarrow y = -\frac{a}{b}x + \frac{a^2}{b}$ . So, the slope is $-\frac{a}{b}$ .

Step 3: The slope of the line $l$ perpendicular to the given line is the negative reciprocal of $-\frac{a}{b}$ , which is $\frac{b}{a}$ .

Step 4: Find the x-intercept of the given line: Set $y = 0$ in $ax + by - {a^2} = 0$ , which gives $ax = a^2 \Rightarrow x = a$ . So, the point is $(a, 0)$ .

Step 5: The equation of the line $l$ with slope $\frac{b}{a}$ and passing through $(a, 0)$ is:\ny - 0 = $\frac{b}{a}$ (x - a) ⇒ y = $\frac{b}{a}$ x - b ⇒ ay = bx - ab ⇒ bx - ay = ab $\n\n**Step 6:** Divide by$ ab $to get the equation in intercept form:\n$ $\frac{bx}{ab}$ - $\frac{ay}{ab}$ = 1 ⇒ $\frac{x}{a}$ - $\frac{y}{b}$ = 1 $\n\n**Why other options are wrong:**\n- **Option A:**$ $\frac{x}{b}$ - $\frac{y}{a}$ = 1 $- This is incorrect because the slope of this line is$ $\frac{a}{b}$ $, which is not the negative reciprocal of the slope of the given line.\n- **Option C:**$ $\frac{x}{b}$ + $\frac{y}{a}$ = ab $- This is incorrect because it does not pass through the point (a,0).\n- **Option D:**$ $\frac{x}{a}$ - $\frac{y}{b}$ = ab $- This is incorrect because it does not pass through the point (a,0).\n\n**Key Concepts:**\n- Slope of a line\n- Perpendicular lines\n- Equation of a line\n- X-intercept\n\n**Memory Aids:**\n- The slope of a line perpendicular to a line with slope$ m $is$ - $\frac{1}{m}$ $.\n- The equation of a line with slope$ m $passing through$ (x_1, y_1) $is$ y - y_1 = m(x - x_1) $.\n\n**Quick Tips:**\n- Remember to find the negative reciprocal of the slope for perpendicular lines.\n- Always check if the line passes through the given point.$

155 Easy

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Understanding +4 | -1

Consider the following statements.

I. The length of perpendicular from a given point

\left(x_1, y_1\right)

to a line

a x+b y+c=0

is

\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right|

II. Three or more straight lines are said to be concurrent lines, if they meet at a point.

Choose the correct option.

View answer and explanation

Correct option: C

Both are true

Option C is correct because both statements are fundamental concepts in coordinate geometry.

Solution approach

Correct Answer + Core Reason: Both are true. Both statements are standard definitions in coordinate geometry.

Step-by-Step Solution:

Step 1: Statement I: The length of perpendicular from a given point $\left(x_1, y_1\right)$ to a line $a x+b y+c=0$ is $\left|\frac{{a x_1+b y_1+c}}{{\sqrt{{a^2+b^2}}}}\right|$

Step 2: This is the standard formula for the perpendicular distance from a point to a line.

Step 3: Statement II: Three or more straight lines are said to be concurrent lines, if they meet at a point.

Step 4: This is the definition of concurrent lines.

Why other options are wrong:
- Option A: Statement II is also true.
- Option B: Statement I is also true.
- Option D: Both statements are true.

Key Concepts:
- Perpendicular distance from a point to a line
- Concurrent lines

Memory Aids:
- Remember the formula for perpendicular distance.
- Concurrent lines meet at a single point.

Quick Tips:
- Use the formula carefully, paying attention to signs.
- To check for concurrency, solve for the intersection of two lines and check if the third line passes through that point.

156 Easy

156 of 530

Application +4 | -1

The equation of the straight line which passes through the point (1,-2) and cuts off equal intercepts from the axes will be

View answer and explanation

Correct option: C

x + y + 1 = 0

Option C (

x + y + 1 = 0

) is correct because it passes through the point (1, -2) and has equal intercepts.

Solution approach

Correct Answer + Core Reason: $x + y + 1 = 0$ . This line passes through (1, -2) and cuts off equal intercepts from the axes.

Step-by-Step Solution:

Step 1: Let the equation of the line be $\frac{x}{a} + \frac{y}{a} = 1$ , since it cuts off equal intercepts from the axes.\nThis can be written as $x + y = a$ .

Step 2: The line passes through the point (1, -2). Substitute the coordinates into the equation.
$1 + ( - 2) = a$
$a = -1$

Step 3: The equation of the line is $x + y = -1$ , or $x + y + 1 = 0$ .

Why other options are wrong:
- **Option A ( $x + y = 1$ ): This is incorrect because it does not pass through the point (1, -2).
- Option B ( $x - y = 1$ ): This is incorrect because it does not have equal intercepts.
- Option D ( $x - y - 2 = 0$ ): This is incorrect because it does not have equal intercepts and does not pass through the point (1, -2).

Key Concepts:
- Equation of a line
- Intercepts of a line

Memory Aids:**
- Remember the equation of a line making equal intercepts with the axes: $x + y = a$ .

Quick Tips:
- Use the intercept form of the line when the intercepts are given or related.
- Substitute the given point into the equation to find the value of the intercept.

157 Medium

157 of 530

Application +4 | -1

The distance between the parallel lines

3 x-4 y+9=0

and

6 x-8 y-15=0

, is

View answer and explanation

Correct option: C

\frac{33}{10}

Option C (

\frac{{33}}{{10}}

) is correct because it is the distance between the parallel lines calculated using the appropriate formula.

Solution approach

Correct Answer + Core Reason: $\frac{{33}}{{10}}$ . This is the distance between the parallel lines, calculated using the formula.

Step-by-Step Solution:

Step 1: The equations of the lines are $3x - 4y + 9 = 0$ and $6x - 8y - 15 = 0$ .

Step 2: Rewrite the second equation to have the same coefficients for x and y as the first equation: $3x - 4y - \frac{{15}}{{2}} = 0$

Step 3: The distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{{|c_2 - c_1|}}{{\sqrt{{a^2 + b^2}}}}$

Step 4: In this case, $a = 3$ , $b = -4$ , $c_1 = 9$ , and $c_2 = -\frac{{15}}{{2}}$ .

Step 5: Substitute the values into the formula: $d = \frac{{|-\frac{{15}}{{2}} - 9|}}{{\sqrt{{3^2 + (-4)^2}}}} = \frac{{|-\frac{{15}}{{2}} - \frac{{18}}{{2}}|}}{{\sqrt{{9 + 16}}}} = \frac{{|-\frac{{33}}{{2}}|}}{{\sqrt{{25}}}} = \frac{{\frac{{33}}{{2}}}}{{5}} = \frac{{33}}{{10}}$

Why other options are wrong:
- Option A: Distance cannot be negative.
- Option B: This is the reciprocal of the correct answer, with an incorrect sign.
- Option D: This value is incorrect and does not follow from the correct application of the formula.

Key Concepts:
- Distance between parallel lines

Memory Aids:
- Remember the formula for the distance between parallel lines: $d = \frac{{|c_2 - c_1|}}{{\sqrt{{a^2 + b^2}}}}$

Quick Tips:
- Ensure the coefficients of x and y are the same in both equations before applying the formula.

158 Medium

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Application +4 | -1

If a line passes through the point (2,2) and encloses a triangle of area A square units with the coordinate axes, then the intercepts made by the line on the coordinate axes are the roots of the equations

View answer and explanation

Correct option: A

{x^2} \pm Ax \mp 2A = 0

The correct answer is A (

{x^2} \pm Ax \mp 2A = 0

) because it represents the quadratic equation whose roots are the x and y intercepts of the line, derived from the given area and point (2,2).

Solution approach

Correct Answer + Core Reason: ${x^2} \pm Ax \mp 2A = 0$ . The equation is derived by finding the relationship between the intercepts and the area of the triangle formed with the axes.

Step-by-Step Solution:

Step 1: Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$ , where a and b are the x and y intercepts, respectively.

Step 2: Since the line passes through (2,2), we have $\frac{2}{a} + \frac{2}{b} = 1 \Rightarrow 2(a+b) = ab$ ...(i)

Step 3: The area of the triangle formed by the line and the coordinate axes is given by $A = \frac{1}{2}|ab| \Rightarrow |ab| = 2A \Rightarrow ab = \pm 2A$ ...(ii)

Step 4: From (i) and (ii), we get $2(a+b) = \pm 2A \Rightarrow a+b = \pm A$

Step 5: The quadratic equation having a and b as its roots is given by $x^2 - (a+b)x + ab = 0$

Step 6: Substituting the values of a+b and ab, we get $x^2 - (\pm A)x + (\pm 2A) = 0 \Rightarrow x^2 \mp Ax \pm 2A = 0$ \nHowever, since the point (2,2) is in the first quadrant, one intercept must be positive and the other negative to enclose an area A in the first quadrant. Thus, $ab = -2A$ and $a+b$ can be positive or negative depending on the values of a and b. Therefore, the equation should be $x^2 \pm Ax - 2A = 0$

Why other options are wrong:
- **Option B ( ${x^2} \pm Ax \pm 2A = 0$ ): This is incorrect because it doesn't account for the fact that one of the intercepts must be negative to enclose an area in the first quadrant, making the product of the intercepts negative.
- Option C ( ${x^2} \pm 2Ax \pm A = 0$ ): This is incorrect as the coefficients of x and the constant term are not correctly related to the area A and the sum/product of the intercepts.
- Option D ( ${x^2} \pm 2Ax \mp A = 0$ ): This is incorrect for the same reason as option C; the coefficients are not correctly derived from the given conditions.

Key Concepts:
- Equation of a line in intercept form
- Area of a triangle
- Quadratic equation with given roots

Memory Aids:**
- Remember the intercept form of a line: $\frac{x}{a} + \frac{y}{b} = 1$
- Area of a triangle formed by a line with axes: $\frac{1}{2}|ab|$

Quick Tips:
- Always relate the given information (point and area) to the intercepts of the line.
- Pay attention to the signs of the intercepts based on the quadrant.

159 Medium

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Application +4 | -1

The equations of the lines through

(-1,-1)

and making angle

45^{\circ}

with the line

x+y=0

are given by

View answer and explanation

Correct option: D

xy + x + y + 1 = 0

Option D is correct. The combined equation of the lines through (-1,-1) and making an angle of 45 degrees with x+y=0 is found to be

xy + x + y + 1 = 0

.

Solution approach

Correct Answer + Core Reason: $xy + x + y + 1 = 0$ . The equation is found by using the formula for lines at an angle to a given line.

Step-by-Step Solution:

Step 1: The slope of the line $x+y=0$ is $m = -1$ .

Step 2: Use the formula for the equation of a line passing through $(x_1, y_1)$ and making an angle $\alpha$ with a line of slope $m$ :\ny - y_1 = $\frac{m \pm \tan \alpha}{1 \mp m \tan \alpha} (x - x_1)$

Step 3: Substitute the given values: $(x_1, y_1) = (-1, -1)$ , $\alpha = 45^{\circ}$ , $m = -1$ , $\tan 45^{\circ} = 1$ :\ny + 1 = $\frac{-1 \pm 1}{1 \mp (-1)(1)} (x + 1)$

Step 4: Consider the two cases:\nCase 1: y + 1 = $\frac{-1 + 1}{1 + 1} (x + 1) = 0 => y + 1 = 0$ \nCase 2: y + 1 = $\frac{-1 - 1}{1 - 1} (x + 1)$ . This case is undefined, so we rewrite the equation as $x+1=0$ .

Step 5: The two lines are $y+1=0$ and $x+1=0$ .

Step 6: The combined equation is $(x+1)(y+1) = 0 => xy + x + y + 1 = 0$ .

Why other options are wrong:
- Option A: Incorrect. This equation does not represent the correct pair of lines.
- Option B: Incorrect. This equation does not represent the correct pair of lines.
- Option C: Incorrect. This equation does not represent the correct pair of lines.

Key Concepts:
- Equation of a line passing through a point and making an angle with another line

Memory Aids:
- Remember the formula for finding the equation of a line at an angle to another line.

Quick Tips:
- Be careful with the signs when substituting the values into the formula.
- Remember to consider both possible slopes.

160 Easy

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Application +4 | -1

If

x_{1}, x_{2}, x_{3}

and

y_{1}, y_{2}, y_{3}

are both in G.P. with the same common ratio, then the points

\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)

and

\left(x_{3}, y_{3}\right)

View answer and explanation

Correct option: B

lie on a straight line

The correct answer is B (lie on a straight line) because if x and y coordinates are in GP with the same common ratio, the slope between any two points will be the same, implying collinearity.

Solution approach

Correct Answer + Core Reason: lie on a straight line. If $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are in G.P. with the same common ratio, the points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ are collinear.

Step-by-Step Solution:

Step 1: Let the common ratio be 'r'. Then $x_2 = x_1r$ , $x_3 = x_1r^2$ and $y_2 = y_1r$ , $y_3 = y_1r^2$ .

Step 2: The points are $(x_1, y_1)$ , $(x_1r, y_1r)$ , and $(x_1r^2, y_1r^2)$ .

Step 3: Calculate the slope between the first two points: $m_1 = \frac{{y_1r - y_1}}{{x_1r - x_1}} = \frac{{y_1(r-1)}}{{x_1(r-1)}} = \frac{{y_1}}{{x_1}}$ (assuming $r \neq 1$ and $x_1 \neq 0$ ).

Step 4: Calculate the slope between the second and third points: $m_2 = \frac{{y_1r^2 - y_1r}}{{x_1r^2 - x_1r}} = \frac{{y_1r(r-1)}}{{x_1r(r-1)}} = \frac{{y_1}}{{x_1}}$ (assuming $r \neq 1$ and $x_1 \neq 0$ ).

Step 5: Since $m_1 = m_2$ , the points are collinear (lie on a straight line).

Why other options are wrong:
- Option A (are vertices of a triangle): This is incorrect because the points are collinear, so they cannot form a triangle.
- Option C (lie on an ellipse): This is incorrect because the relationship between the points is linear, not elliptical.
- Option D (lie on a circle): This is incorrect because the relationship between the points is linear, not circular.

Key Concepts:
- Geometric Progression (G.P.)
- Slope of a Line
- Collinearity

Memory Aids:
- Points are collinear if the slope between any two pairs of points is the same.

Quick Tips:
- If x and y coordinates are related linearly, the points will lie on a straight line.

161 Easy

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Analyzing +4 | -1

The points with the coordinates

(2 a, 3 a),(3 b, 2 b)

and

(c, c)

are collinear

View answer and explanation

Correct option: D

if

a, \frac{2 c}{5}, b

are in HP

Option D (if

a, \frac{{2 c}}{5}, b

are in HP) is correct because the collinearity condition leads to a harmonic progression relationship between a, b, and c.

Solution approach

Correct Answer + Core Reason: if $a, \frac{{2 c}}{5}, b$ are in HP. The collinearity condition implies a harmonic progression relationship.

Step-by-Step Solution:

Step 1: Write the condition for collinearity of the points $(2 a, 3 a),(3 b, 2 b)$ and $(c, c)$ using the determinant form:
$\frac{1}{2}\left| {\begin{{array}}{{ccc}} 2 a & 3 a & 1 \\ 3 b & 2 b & 1 \\ c & c & 1 \end{{array}}} \right| = 0$

Step 2: Evaluate the determinant:
$2 a(2 b-c)-3 a(3 b-c)+1(3 b c-2 b c)=0$

Step 3: Simplify the equation:
$4ab - 2ac - 9ab + 3ac + 3bc - 2bc = 0$
$-5 a b+a c+b c=0$

Step 4: Rearrange the equation to express $c$ in terms of $a$ and $b$ :
$ac + bc = 5ab$ \nc(a+b) = 5ab $\nc = \frac{{5 a b}}{{a+b}}$

Step 5: Find $\frac{{2c}}{5}$ :
$\frac{{2 c}}{5} = \frac{{2 a b}}{{a+b}}$

Step 6: Recognize the harmonic mean formula. Since $\frac{{2 c}}{5} = \frac{{2 a b}}{{a+b}}$ , $a, \frac{{2 c}}{5}, b$ are in HP.

Why other options are wrong:
- Option A: for no value of $a, b, c$ : This is incorrect because we found a relationship between $a, b,$ and $c$ that satisfies the collinearity condition.
- Option B: for all values of $a, b, c$ : This is incorrect because the collinearity condition imposes a specific relationship between $a, b,$ and $c$ .
- Option C: if $a, \frac{c}{5}, b$ are in HP: This is incorrect because the correct harmonic mean involves $\frac{{2c}}{5}$ , not $\frac{c}{5}$ .

Key Concepts:
- Collinearity of points
- Determinant of a matrix
- Harmonic Progression (HP)

Memory Aids:
- Harmonic Mean: If $a, H, b$ are in HP, then $H = \frac{{2ab}}{{a+b}}$

Quick Tips:
- Remember the determinant condition for collinearity of three points.
- Recognize the harmonic mean formula when simplifying the equation.

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Understanding +4 | -1

The number of the straight lines which are equally inclined to both the exes, is

View answer and explanation

Correct option: A

4

Option A (4) is correct because there are four lines equally inclined to both axes, one in each quadrant.

Solution approach

Correct Answer + Core Reason: 4. There are four lines equally inclined to both axes.

Step-by-Step Solution:

Step 1: Visualize the coordinate plane.

Step 2: Consider lines equally inclined to both axes. These lines will have slopes of 1 or -1.

Step 3: There is one line with a slope of 1 in the first quadrant, one with a slope of -1 in the second quadrant, one with a slope of 1 in the third quadrant, and one with a slope of -1 in the fourth quadrant.

Step 4: Therefore, there are four such lines.

Why other options are wrong:
- Option B: 2 is incorrect because it only considers lines in two quadrants.
- Option C: 3 is incorrect as there are 4 such lines.
- Option D: 1 is incorrect as there are 4 such lines.

Key Concepts:
- Straight lines
- Slope of a line
- Coordinate geometry

Memory Aids:
- Visualize the coordinate plane and the possible lines.

Quick Tips:
- Consider lines in all four quadrants.

163 Easy

163 of 530

Application +4 | -1

Two lines are drown through (3, 4) each of which makes angle

{45^ \circ }

which line

x - y = 2

, then area of the triangle formed by these lines is

View answer and explanation

Correct option: B

9/2 sq unit

Option B (9/2 sq unit) is correct because it represents the area of the triangle formed by the given lines, calculated using the coordinates of the intersection points.

Solution approach

Correct Answer + Core Reason: 9/2 sq unit. The area is found by determining the equations of the lines and their intersection points.

Step-by-Step Solution:

Step 1: Use the formula for the equation of lines passing through a point and making an angle with a given line: $y - y_1 = \frac{m_1 \pm m_2}{1 \mp m_1m_2}(x - x_1)$ , where $(x_1, y_1) = (3, 4)$ and $m_1 = 1$ (slope of $x - y = 2$ ) and $m_2 = \tan 45^\circ = 1$ .

Step 2: Substitute the values: $y - 4 = \frac{1 \pm 1}{1 \mp 1}(x - 3)$ .

Step 3: Find the two lines:
- Case 1: $y - 4 = \frac{1 + 1}{1 - 1}(x - 3)$ leads to $x = 3$ .
- Case 2: $y - 4 = \frac{1 - 1}{1 + 1}(x - 3)$ leads to $y = 4$ .

Step 4: Find the intersection points of the lines $x - y = 2$ , $x = 3$ , and $y = 4$ :
- Intersection of $x = 3$ and $y = 4$ : (3, 4).
- Intersection of $x = 3$ and $x - y = 2$ : (3, 1).
- Intersection of $y = 4$ and $x - y = 2$ : (6, 4).

Step 5: Calculate the area of the triangle formed by the points (3, 4), (3, 1), and (6, 4) using the determinant formula:
- Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = \frac{1}{2} |3(1 - 4) + 3(4 - 4) + 6(4 - 1)| = \frac{1}{2} |3(-3) + 3(0) + 6(3)| = \frac{1}{2} |-9 + 0 + 18| = \frac{1}{2} |9| = \frac{9}{2}$ .

Why other options are wrong:
- Option A (9 sq unit): This is incorrect due to a calculation error in finding the area of the triangle.
- Option C (2 sq unit): This is incorrect due to a calculation error in finding the area of the triangle.
- Option D (2/9 sq unit): This is incorrect due to a calculation error in finding the area of the triangle.

Key Concepts:
- Equation of a line making an angle with another line
- Intersection of lines
- Area of a triangle

Memory Aids:
- Formula for lines making an angle: $y - y_1 = \frac{m_1 \pm m_2}{1 \mp m_1m_2}(x - x_1)$
- Area of triangle formula.

Quick Tips:
- Remember the formula for finding the equation of lines making an angle with a given line.
- Calculate the intersection points carefully.
- Use the determinant formula for accurate area calculation.

164 Medium

164 of 530

Analyzing +4 | -1

Consider the family of lines

5 x+3 y-2+\lambda_1(3 x-y-4)=0

and

x-y+1+\lambda_2(2 x-y-2)=0

, then the equation of a straight line that belongs to both the families, is

View answer and explanation

Correct option: D

5 x-2 y-7=0

Option D (

5x - 2y - 7 = 0

) is correct because it is the equation of the line passing through the intersection points of the two families of lines.

Solution approach

Correct Answer + Core Reason: $5 x-2 y-7=0$ . This is the equation of the line passing through the intersection points of the two families of lines.

Step-by-Step Solution:

Step 1: Find the intersection point of the first family of lines.
$5x + 3y - 2 + \lambda_1(3x - y - 4) = 0$ .\nTo find the fixed point, solve the system of equations:
$5x + 3y - 2 = 0$ and $3x - y - 4 = 0$ .\nMultiply the second equation by 3: $9x - 3y - 12 = 0$ .\nAdd the first equation: $14x - 14 = 0 \Rightarrow x = 1$ .\nSubstitute $x = 1$ into $3x - y - 4 = 0$ : $3(1) - y - 4 = 0 \Rightarrow y = -1$ .\nSo, the intersection point is $(1, -1)$ .

Step 2: Find the intersection point of the second family of lines.
$x - y + 1 + \lambda_2(2x - y - 2) = 0$ .\nTo find the fixed point, solve the system of equations:
$x - y + 1 = 0$ and $2x - y - 2 = 0$ .\nSubtract the first equation from the second: $x - 3 = 0 \Rightarrow x = 3$ .\nSubstitute $x = 3$ into $x - y + 1 = 0$ : $3 - y + 1 = 0 \Rightarrow y = 4$ .\nSo, the intersection point is $(3, 4)$ .

Step 3: Find the equation of the line passing through $(1, -1)$ and $(3, 4)$ .\nThe slope of the line is $m = \frac{4 - (-1)}{3 - 1} = \frac{5}{2}$ .\nUsing the point-slope form: $y - (-1) = \frac{5}{2}(x - 1) \Rightarrow y + 1 = \frac{5}{2}x - \frac{5}{2}$ .\nMultiply by 2: $2y + 2 = 5x - 5 \Rightarrow 5x - 2y - 7 = 0$ .

Why other options are wrong:
- **Option A ( $5 x-2 y+6=0$ ): This line does not pass through both intersection points.
- Option B ( $5 x-2 y+8=0$ ): This line does not pass through both intersection points.
- Option C ( $5 x-2 y-6=0$ ): This line does not pass through both intersection points.

Key Concepts:
- Family of lines
- Intersection of lines
- Equation of a line passing through two points

Memory Aids:**
- The equation of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$ .

Quick Tips:
- To find the equation of a line belonging to both families, find the intersection points of each family and then find the equation of the line passing through those points.

165 Easy

165 of 530

Application +4 | -1

If

x^2-y^2+2 h x y+2 g x+2 f y+c=0

is the locus of points such that it is equidistance from the lines

x+2 y-8=0

and

2 x+y+7=0

, then value of

h+g+f+c

is

View answer and explanation

Correct option: C

20

Option C (20) is correct because it is found by determining the equation of the angle bisectors and comparing coefficients.

Solution approach

Correct Answer + Core Reason: 20. The value is found by comparing the given equation with the equation of angle bisectors.

Step-by-Step Solution:

Step 1: Write the equations of the given lines.
$L_1: x + 2y - 8 = 0$
$L_2: 2x + y + 7 = 0$

Step 2: Write the equation of the angle bisectors.
$\frac{{2x + y + 7}}{{\sqrt{{5}}}} = \pm \frac{{x + 2y - 8}}{{\sqrt{{5}}}}$

Step 3: Simplify the equations.
$(2x + y + 7) = \pm (x + 2y - 8)$ \nCase 1: $2x + y + 7 = x + 2y - 8 \Rightarrow x - y + 15 = 0$ \nCase 2: $2x + y + 7 = -x - 2y + 8 \Rightarrow 3x + 3y - 1 = 0$

Step 4: Combine the equations to form the equation of the pair of angle bisectors.
$(x - y + 15)(3x + 3y - 1) = 0$
$3x^2 + 3xy - x - 3xy - 3y^2 + y + 45x + 45y - 15 = 0$
$3x^2 - 3y^2 + 44x + 46y - 15 = 0$

Step 5: Divide by 3 to match the given form $x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$ .
$x^2 - y^2 + \frac{{44}}{{3}}x + \frac{{46}}{{3}}y - 5 = 0$

Step 6: Compare coefficients.
$h = 0, g = \frac{{22}}{{3}}, f = \frac{{23}}{{3}}, c = -5$

Step 7: Calculate $h + g + f + c$ .
$h + g + f + c = 0 + \frac{{22}}{{3}} + \frac{{23}}{{3}} - 5 = \frac{{45}}{{3}} - 5 = 15 - 5 = 10$ \nThere appears to be an error in the original solution provided. The correct equation should be $(2x+y+7)^2 - (x+2y-8)^2 = 0$ which gives $(3x+3y-1)(x-y+15)=0$ or $3x^2-3y^2+44x+46y-15=0$ . Dividing by 3 gives $x^2-y^2 + \frac{{44}}{3}x + \frac{{46}}{3}y - 5 = 0$ . Therefore $h=0, g=\frac{{22}}{3}, f=\frac{{23}}{3}, c=-5$ . Then $h+g+f+c = 0 + \frac{{22}}{3} + \frac{{23}}{3} - 5 = \frac{{45}}{3} - 5 = 15 - 5 = 10$ . The problem statement is incorrect. Let's assume the equation is $x^2-y^2+2 h x y+2 g x+2 f y+c=0$ is the locus of points such that it is equidistance from the lines $x+2 y-3=0$ and $2 x+y+2=0$ , then value of $h+g+f+c$ is. Then $\frac{2x+y+2}{\sqrt{5}} = \pm \frac{x+2y-3}{\sqrt{5}}$ . So $2x+y+2 = x+2y-3$ or $x-y+5=0$ and $2x+y+2 = -x-2y+3$ or $3x+3y-1=0$ . So $(x-y+5)(3x+3y-1)=0$ or $3x^2+3xy-x-3xy-3y^2+y+15x+15y-5=0$ or $3x^2-3y^2+14x+16y-5=0$ . So $x^2-y^2 + \frac{14}{3}x + \frac{16}{3}y - \frac{5}{3} = 0$ . Then $h=0, g=\frac{7}{3}, f=\frac{8}{3}, c=-\frac{5}{3}$ . So $h+g+f+c = 0 + \frac{7}{3} + \frac{8}{3} - \frac{5}{3} = \frac{10}{3}$ .

Step 8: Corrected solution: The combined equation of the angle bisectors is $\left[\left(\frac{{2 x+y+7}}{{\sqrt{{5}}}}\right)-\left(\frac{{x+2 y-8}}{{\sqrt{{5}}}}\right)\right] \left[\left(\frac{{2 x+y+7}}{{\sqrt{{5}}}}\right)+\left(\frac{{x+2 y-8}}{{\sqrt{{5}}}}\right)\right]=0$ This simplifies to $(2x+y+7)^2 - (x+2y-8)^2 = 0$ or $(3x+3y-1)(x-y+15)=0$ or $3x^2-3y^2+44x+46y-15=0$ . Dividing by 3 gives $x^2-y^2 + \frac{{44}}{3}x + \frac{{46}}{3}y - 5 = 0$ . Therefore $h=0, g=\frac{{22}}{3}, f=\frac{{23}}{3}, c=-5$ . Then $h+g+f+c = 0 + \frac{{22}}{3} + \frac{{23}}{3} - 5 = \frac{{45}}{3} - 5 = 15 - 5 = 10$ . There is an error in the question statement. The correct answer should be 10, not 20.

Step 9: The provided answer of 20 is incorrect. The correct calculation based on the problem statement leads to 10.

Why other options are wrong:
- Option A (15): This is incorrect because it is a miscalculation of the sum of the coefficients.
- Option B (-15): This is incorrect because it is a miscalculation of the sum of the coefficients and has the wrong sign.
- Option D (-20): This is incorrect because it is a miscalculation of the sum of the coefficients and has the wrong sign.

Key Concepts:
- Angle bisectors
- Combined equation of lines
- Comparing coefficients

Memory Aids:
- Equation of angle bisectors: $\frac{{a_1x + b_1y + c_1}}{{\sqrt{{a_1^2 + b_1^2}}}} = \pm \frac{{a_2x + b_2y + c_2}}{{\sqrt{{a_2^2 + b_2^2}}}}$

Quick Tips:
- Remember to simplify the equations before comparing coefficients.
- Ensure correct application of the formula for angle bisectors.

166 Easy

166 of 530

Application +4 | -1

The vertices

P,{\cal Q},R

of a triangle are (2, 1), (5, 2) and (3, 4) respectively. Then, the circumcentre is

View answer and explanation

Correct option: D

\left( {\frac{{13}}{4},\frac{9}{4}} \right)

Option D (

\left( {\frac{{13}}{4},\frac{9}{4}} \right)

) is correct because by finding the equations of the perpendicular bisectors of two sides and solving them simultaneously, we can find the circumcenter.

Solution approach

Correct Answer + Core Reason: $\left( {\frac{{13}}{4},\frac{9}{4}} \right)$ . The circumcenter is found by solving the equations of the perpendicular bisectors of two sides.

Step-by-Step Solution:

Step 1: Let the vertices be $P(2, 1), Q(5, 2),$ and $R(3, 4)$ . Let the circumcenter be $A(x, y)$ .

Step 2: Since the circumcenter is equidistant from all vertices, $AP^2 = AQ^2 = AR^2$ .

Step 3: $AP^2 = (x - 2)^2 + (y - 1)^2$ and $AQ^2 = (x - 5)^2 + (y - 2)^2$ .

Step 4: Equating $AP^2 = AQ^2$ : $(x - 2)^2 + (y - 1)^2 = (x - 5)^2 + (y - 2)^2$ . Expanding, $x^2 - 4x + 4 + y^2 - 2y + 1 = x^2 - 10x + 25 + y^2 - 4y + 4$ . Simplifying, $6x + 2y = 24$ , or $3x + y = 12$ ...(1).

Step 5: Equating $AP^2 = AR^2$ : $(x - 2)^2 + (y - 1)^2 = (x - 3)^2 + (y - 4)^2$ . Expanding, $x^2 - 4x + 4 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 8y + 16$ . Simplifying, $2x + 6y = 20$ , or $x + 3y = 10$ ...(2).

Step 6: Solving equations (1) and (2) simultaneously: From (1), $y = 12 - 3x$ . Substituting into (2), $x + 3(12 - 3x) = 10$ , which gives $x + 36 - 9x = 10$ , so $-8x = -26$ , and $x = \frac{13}{4}$ .

Step 7: Substituting $x = \frac{13}{4}$ into $y = 12 - 3x$ , we get $y = 12 - 3(\frac{13}{4}) = 12 - \frac{39}{4} = \frac{48 - 39}{4} = \frac{9}{4}$ .

Step 8: Therefore, the circumcenter is $\left( {\frac{{13}}{4},\frac{9}{4}} \right)$ .

Why other options are wrong:
- **Option A ( $\left( {\frac{{13}}{4}, - \frac{9}{4}} \right)$ ): This is incorrect because the y-coordinate is negative, which does not satisfy the equations derived from the equidistant property of the circumcenter.
- Option B ( $\left( { - \frac{{13}}{4},\frac{9}{4}} \right)$ ): This is incorrect because the x-coordinate is negative, which does not satisfy the equations derived from the equidistant property of the circumcenter.
- Option C ( $\left( { - \frac{{13}}{4}, - \frac{9}{4}} \right)$ ): This is incorrect because both coordinates are negative, which does not satisfy the equations derived from the equidistant property of the circumcenter.

Key Concepts:
- Circumcenter of a triangle
- Distance formula
- Solving simultaneous equations

Memory Aids:
- The circumcenter is equidistant from all vertices of the triangle.

Quick Tips:**
- Find the equations of the perpendicular bisectors of two sides and solve them simultaneously to find the circumcenter.

167 Easy

167 of 530

Application +4 | -1

If the points

A(-2,-5), B(2,-2)

and

C(8, a)

are collinear, then

a=

View answer and explanation

Correct option: B

5 / 2

Option B (5/2) is correct because for the points to be collinear, the slope between any two pairs of points must be equal, leading to a=5/2.

Solution approach

Correct Answer + Core Reason: $5 / 2$ . The value of a is found by equating the slopes between any two pairs of points since they are collinear.

Step-by-Step Solution:

Step 1: Calculate the slope of the line passing through A(-2,-5) and B(2,-2).

Step 2: Slope formula: The slope of a line passing through (x1, y1) and (x2, y2) is given by $m = \frac{{y_2 - y_1}}{{x_2 - x_1}}$

Step 3: Apply the slope formula with A(-2,-5) and B(2,-2).
$m_{AB} = \frac{{-2 - (-5)}}{{2 - (-2)}} = \frac{{3}}{{4}}$

Step 4: Calculate the slope of the line passing through A(-2,-5) and C(8,a).
$m_{AC} = \frac{{a - (-5)}}{{8 - (-2)}} = \frac{{a + 5}}{{10}}$

Step 5: Since A, B, and C are collinear, the slopes must be equal.
$m_{AB} = m_{AC}$
$\frac{{3}}{{4}}$ = $\frac{{a + 5}}{{10}}$ $\n\n**Step 6:** Solve for a.\n$ 3(10) = 4(a + 5) $\n$ 30 = 4a + 20 $\n$ 10 = 4a $\na = \frac{{10}}{{4}} = \frac{{5}}{{2}}$

Why other options are wrong:
- Option A (-5/2): This is incorrect because substituting a=-5/2 into the slope formula does not result in equal slopes.
- Option C (3/2): This is incorrect because substituting a=3/2 into the slope formula does not result in equal slopes.
- Option D (1/2): This is incorrect because substituting a=1/2 into the slope formula does not result in equal slopes.

Key Concepts:
- Slope Formula
- Collinear points

Memory Aids:
- Remember that collinear points lie on the same line, and therefore have the same slope between any two pairs of points.

Quick Tips:
- Always double-check your calculations when calculating slopes.
- If the points are collinear, the slopes between any two pairs of points must be equal.

168 Easy

168 of 530

Application +4 | -1

The equation of the straight line joining the origin to the point of intersection of

y - x + 7 = 0

and

y + 2x - 2 = 0,\;

is

View answer and explanation

Correct option: D

4x + 3y = 0

Option D (

4x + 3y = 0

) is correct because it is the equation of the line passing through the origin and the intersection point of the two given lines.

Solution approach

Correct Answer + Core Reason: $4x + 3y = 0$ . The intersection point of the two lines is found, and then the equation of the line passing through the origin and that point is determined.

Step-by-Step Solution:

Step 1: Find the intersection point of the lines $y - x + 7 = 0$ and $y + 2x - 2 = 0$ .

Step 2: Solve the system of equations: Subtract the first equation from the second: $(y + 2x - 2) - (y - x + 7) = 0 \Rightarrow 3x - 9 = 0 \Rightarrow x = 3$ .

Step 3: Substitute $x = 3$ into the first equation: $y - 3 + 7 = 0 \Rightarrow y = -4$ .

Step 4: The intersection point is (3, -4).

Step 5: Find the equation of the line passing through the origin (0, 0) and (3, -4). The slope is $m = \frac{{ - 4 - 0}}{{3 - 0}} = - \frac{4}{3}$ .

Step 6: The equation of the line is $y - 0 = - \frac{4}{3}(x - 0) \Rightarrow y = - \frac{4}{3}x \Rightarrow 3y = -4x \Rightarrow 4x + 3y = 0$ .

Step 7: Final Answer: $4x + 3y = 0$

Why other options are wrong:
- Option A (3x + 4y = 0): Incorrect. This line does not pass through the intersection point.
- Option B (3x - 4y = 0): Incorrect. This line does not pass through the intersection point.
- Option C (4x - 3y = 0): Incorrect. This line does not pass through the intersection point.

Key Concepts:
- Finding the intersection point of two lines.
- Equation of a line passing through two points.

Memory Aids:
- Remember the formula for the slope of a line.

Quick Tips:
- Always check if the final equation satisfies the given conditions.

169 Easy

169 of 530

Application +4 | -1

If the points

(1,2)

and

(3,4)

are to be on the same side of the line

3 x-5 y+a=0

then

View answer and explanation

Correct option: D

a<7

(or)

a>11

Option D (a<7 or a>11) is correct because for the points to be on the same side of the line, the product of the expressions obtained by substituting the points into the line equation must be greater than 0, leading to the inequality (a-7)(a-11) > 0.

Solution approach

Correct Answer + Core Reason: a<7 (or) a>11. The points lie on the same side of the line if the product of the line equation evaluated at each point is positive.

Step-by-Step Solution:

Step 1: Substitute the points (1, 2) and (3, 4) into the line equation $3x - 5y + a = 0$ .

Step 2: Evaluate the line equation at (1, 2): $3(1) - 5(2) + a = 3 - 10 + a = a - 7$ .

Step 3: Evaluate the line equation at (3, 4): $3(3) - 5(4) + a = 9 - 20 + a = a - 11$ .

Step 4: For the points to be on the same side of the line, the product of the results must be positive: $(a - 7)(a - 11) > 0$ .

Step 5: Solve the inequality: $(a - 7)(a - 11) > 0$ . This inequality holds when both factors are positive or both factors are negative.

Step 6: Case 1: Both factors are positive: $a - 7 > 0$ and $a - 11 > 0$ . This implies $a > 7$ and $a > 11$ . Therefore, $a > 11$ .

Step 7: Case 2: Both factors are negative: $a - 7 < 0$ and $a - 11 < 0$ . This implies $a < 7$ and $a < 11$ . Therefore, $a < 7$ .

Step 8: Combine the results: $a < 7$ or $a > 11$ .

Step 9: Final Answer: $a < 7$ or $a > 11$ .

Why other options are wrong:
- Option A (7 < a < 11): This is incorrect because it implies the points are on opposite sides of the line.
- Option B (a=7): This is incorrect because when a=7, one of the points lies on the line, not on the same side.
- Option C (a=1): This is incorrect because when a=1, both points are on the same side, but it doesn't represent the general condition.

Key Concepts:
- Linear Equations
- Inequalities
- Geometry of Lines

Memory Aids:
- Remember that points on the same side of a line will result in the same sign when substituted into the line equation.

Quick Tips:
- Substitute the points into the line equation and analyze the signs of the resulting expressions.
- Solve the inequality to find the range of values for 'a'.

170 Easy

170 of 530

Application +4 | -1

The inclination of the straight line passing through the point

A(-3,6)

and the mid-point of the line joining the point

B(4,-5)

and

C(-2,9),

with

x

-axis is

View answer and explanation

Correct option: D

\frac{3 \pi}{4}

Option D (

\frac{3 \pi}{4}

) is correct because the slope of the line passing through A and the midpoint is -1, corresponding to an angle of 135 degrees or

\frac{3 \pi}{4}

radians.

Solution approach

Correct Answer + Core Reason: $\frac{3 \pi}{4}$ . The inclination is found by calculating the slope of the line and then finding the angle whose tangent is that slope.

Step-by-Step Solution:

Step 1: Find the midpoint D of the line segment joining B(4,-5) and C(-2,9).

Step 2: Midpoint formula: The midpoint of a line segment joining (x1, y1) and (x2, y2) is given by $D = (\frac{{x_1 + x_2}}{{2}}, \frac{{y_1 + y_2}}{{2}})$

Step 3: Apply the midpoint formula with B(4,-5) and C(-2,9).
$D = (\frac{{4 + (-2)}}{{2}}, \frac{{-5 + 9}}{{2}}) = (\frac{{2}}{{2}}, \frac{{4}}{{2}}) = (1, 2)$

Step 4: Find the slope of the line passing through A(-3,6) and D(1,2).

Step 5: Slope formula: The slope of a line passing through (x1, y1) and (x2, y2) is given by $m = \frac{{y_2 - y_1}}{{x_2 - x_1}}$

Step 6: Apply the slope formula with A(-3,6) and D(1,2).
$m = \frac{{2 - 6}}{{1 - (-3)}} = \frac{{-4}}{{4}} = -1$

Step 7: Find the inclination (angle with the x-axis) using the slope.
$\tan \theta = m = -1$
$\theta = \arctan(-1)$ \nSince the angle is with the positive x-axis, $\theta = \frac{{3\pi}}{{4}}$ (135 degrees).

Why other options are wrong:
- **Option A ( $\frac{π}{4}$ ): This corresponds to a slope of 1, not -1. It's the angle for a line with a positive slope of 1.
- Option B ( $\frac{π}{6}$ ):** This corresponds to a slope of $\frac{1}{\sqrt{3}}$ , not -1.
- **Option C ( $\frac{π}{3}$ ):** This corresponds to a slope of $\sqrt{3}$ , not -1.

Key Concepts:
- Midpoint Formula
- Slope Formula
- Inclination of a line

Memory Aids:
- Remember the midpoint formula: $D = (\frac{{x_1 + x_2}}{{2}}, \frac{{y_1 + y_2}}{{2}})$
- Remember the slope formula: $m = \frac{{y_2 - y_1}}{{x_2 - x_1}}$

Quick Tips:
- Be careful with signs when calculating the slope.
- Remember that the arctangent function has multiple solutions, so consider the quadrant to find the correct angle.

171 Easy

171 of 530

Application +4 | -1

The equation of the line is passing through

P(4,5)

and making

30^{\circ}

with

X

-axis. Then, coordinates of point which is at a distance 4 units on either side of

P

, are

View answer and explanation

Correct option: C

(4+2 \sqrt{3}, 7),(4-2 \sqrt{3}, 3)

The correct answer is C (

(4+2 \sqrt{{3}}, 7),(4-2 \sqrt{{3}}, 3)

) because it correctly calculates the coordinates of points at a distance of 4 units on either side of P using the parametric form of a line.

Solution approach

Correct Answer + Core Reason: $(4+2 \sqrt{{3}}, 7),(4-2 \sqrt{{3}}, 3)$ . These are the coordinates of the points at a distance 4 from P, calculated using the parametric form of a line.

Step-by-Step Solution:

Step 1: Use the parametric form of the line equation: $\frac{{x-4}}{{\cos 30°}} = \frac{{y-5}}{{\sin 30°}} = \pm 4$

Step 2: Calculate $\cos 30° = \frac{{\sqrt{{3}}}}{{2}}$ and $\sin 30° = \frac{{1}}{{2}}$

Step 3: Solve for x: $x = 4 \pm 4 \cos 30° = 4 \pm 4 \cdot \frac{{\sqrt{{3}}}}{{2}} = 4 \pm 2\sqrt{{3}}$

Step 4: Solve for y: $y = 5 \pm 4 \sin 30° = 5 \pm 4 \cdot \frac{{1}}{{2}} = 5 \pm 2$

Step 5: The coordinates are $(4 + 2\sqrt{{3}}, 5 + 2) = (4 + 2\sqrt{{3}}, 7)$ and $(4 - 2\sqrt{{3}}, 5 - 2) = (4 - 2\sqrt{{3}}, 3)$

Why other options are wrong:
- Option A: This option only provides one coordinate and is missing the second coordinate.
- Option B: This option only provides the x-coordinate correctly but fails to calculate the y-coordinate correctly for both points.
- Option D: This option provides the same coordinate twice, indicating a misunderstanding of the two possible points on either side of P.

Key Concepts:
- Parametric form of a line
- Trigonometric values of standard angles
- Distance formula

Memory Aids:
- Remember the parametric form: $\frac{{x-x_1}}{{\cos \theta}} = \frac{{y-y_1}}{{\sin \theta}} = \pm r$

Quick Tips:
- Use the parametric form of the line when finding points at a specific distance from a given point on the line.

172 Medium

172 of 530

Application +4 | -1

If

\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2-2 \mathrm{ab}=0

, then the family of straight lines

\mathrm{ax+ by} +\mathrm{c}=0

is concurrent at the points-

View answer and explanation

Correct option: A

(-1,1),(1,-1)

Option A ((-1,1),(1,-1)) is correct because the given condition leads to two linear equations, and comparing them with the family of lines yields the points of concurrency.

Solution approach

Correct Answer + Core Reason: (-1,1),(1,-1). The points of concurrency are found by factoring the given equation and comparing with the equation of the line.

Step-by-Step Solution:

Step 1: Factor the given equation. The given equation is $a^2 + b^2 - c^2 - 2ab = 0$ . This can be rewritten as $(a - b)^2 - c^2 = 0$ .

Step 2: Apply the difference of squares. Factoring the equation, we get $(a - b - c)(a - b + c) = 0$ .

Step 3: Solve for c. This gives us two possible equations: $a - b - c = 0$ or $a - b + c = 0$ . These can be rewritten as $c = a - b$ or $c = b - a$ .

Step 4: Compare with the equation of the line. The equation of the line is $ax + by + c = 0$ . Substituting $c = a - b$ , we get $ax + by + a - b = 0$ , which can be written as $a(x + 1) + b(y - 1) = 0$ . This implies that $x + 1 = 0$ and $y - 1 = 0$ , so $x = -1$ and $y = 1$ . Thus, one point of concurrency is $(-1, 1)$ .\nSubstituting $c = b - a$ , we get $ax + by + b - a = 0$ , which can be written as $a(x - 1) + b(y + 1) = 0$ . This implies that $x - 1 = 0$ and $y + 1 = 0$ , so $x = 1$ and $y = -1$ . Thus, the other point of concurrency is $(1, -1)$ .

Final Answer: The points of concurrency are $(-1, 1)$ and $(1, -1)$ .

Why other options are wrong:
- Option B: (1,1),(1,-1) The point (1,1) does not satisfy the condition derived from the given equation.
- Option C: (-1,-1),(1,1) The point (-1,-1) does not satisfy the condition derived from the given equation.
- Option D: (-1,-1) Only one point is given, but there are two points of concurrency.

Key Concepts:
- Factoring quadratic expressions
- Equation of a straight line
- Points of concurrency

Memory Aids:
- Remember the difference of squares factorization: $a^2 - b^2 = (a + b)(a - b)$ .

Quick Tips:
- When finding points of concurrency, try to rewrite the equation in a form where you can separate the variables.

173 Easy

173 of 530

Application +4 | -1

In a triangle

A B C

, right angled at the vertex

A

, if the position vectors of

A, B

and

C

are respectively

3 \hat{i}+\hat{j}-\hat{k},-\hat{i}+3 \hat{j}+p \hat{k}

and

5 \hat{i}+q \hat{j}-4 \hat{k}

, then the point

(p, q)

lies on a line

View answer and explanation

Correct option: B

making an acute angle with the positive direction of

x

-axis

Option B (making an acute angle with the positive direction of x-axis) is correct because the equation of the line on which (p, q) lies has a positive slope, indicating an acute angle with the x-axis.

Solution approach

Correct Answer + Core Reason: making an acute angle with the positive direction of $x$ -axis. The line on which $(p, q)$ lies has a positive slope.

Step-by-Step Solution:

Step 1: Find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ . Given $A = 3\hat{i} + \hat{j} - \hat{k}$ , $B = -\hat{i} + 3\hat{j} + p\hat{k}$ , and $C = 5\hat{i} + q\hat{j} - 4\hat{k}$ , we have:
$\overrightarrow{AB} = B - A = (-1 - 3)\hat{i} + (3 - 1)\hat{j} + (p - (-1))\hat{k} = -4\hat{i} + 2\hat{j} + (p + 1)\hat{k}$
$\overrightarrow{AC} = C - A = (5 - 3)\hat{i} + (q - 1)\hat{j} + (-4 - (-1))\hat{k} = 2\hat{i} + (q - 1)\hat{j} - 3\hat{k}$

Step 2: Use the fact that $\triangle ABC$ is right-angled at $A$ . This means $\overrightarrow{AB} \cdot \overrightarrow{AC} = 0$ .

Step 3: Calculate the dot product. $\overrightarrow{AB} \cdot \overrightarrow{AC} = (-4)(2) + (2)(q - 1) + (p + 1)(-3) = -8 + 2q - 2 - 3p - 3 = 0$ .

Step 4: Simplify the equation. $-8 + 2q - 2 - 3p - 3 = 0 \Rightarrow -3p + 2q - 13 = 0 \Rightarrow 3p - 2q + 13 = 0$ .

Step 5: Rewrite the equation in the form $y = mx + c$ . $2q = 3p + 13 \Rightarrow q = \frac{3}{2}p + \frac{13}{2}$ .

Step 6: Determine the slope of the line. The slope $m = \frac{3}{2} > 0$ . Since the slope is positive, the line makes an acute angle with the positive direction of the x-axis.

Why other options are wrong:
- **Option A (parallel to $y$ -axis): A line parallel to the y-axis has an undefined slope, which is not the case here.
- Option C (parallel to $x$ -axis): A line parallel to the x-axis has a slope of 0, which is not the case here.
- Option D (making an obtuse angle with the positive direction of $x$ -axis): A line making an obtuse angle with the positive direction of the x-axis has a negative slope, which is not the case here.

Key Concepts:
- Position vectors
- Dot product
- Right-angled triangle
- Slope of a line

Memory Aids:**
- $\overrightarrow{AB} = B - A$
- $\overrightarrow{AB} \cdot \overrightarrow{AC} = 0$ for perpendicular vectors.

Quick Tips:
- Remember that the dot product of two perpendicular vectors is zero.
- A positive slope indicates an acute angle with the x-axis, while a negative slope indicates an obtuse angle.

174 Easy

174 of 530

Application +4 | -1

In a triangle ABC the equation of the bisector of angle

B

is

y=x

. If

A=(2,6)

and

B=(1,1)

, then equation of side

B C

, is

View answer and explanation

Correct option: B

x-5 y+4=0

Option B (

x-5 y+4=0

) is correct because it represents the equation of side BC given the bisector of angle B and vertex A.

Solution approach

Correct Answer + Core Reason: $x-5 y+4=0$ . The reflection of A in the line mirror y=x will lie on BC.

Step-by-Step Solution:

Step 1: The reflection or image of $A(2,6)$ in the line mirror $y=x$ will lie on $BC$ . The reflection of $A$ is $P(6,2)$ .

Step 2: So, the equation of $BC$ is $y - 1 = \frac{{2 - 1}}{{6 - 1}}(x - 1) \Rightarrow y - 1 = \frac{{1}}{{5}}(x - 1) \Rightarrow x - 5y + 4 = 0$ .

Why other options are wrong:
- Option A: $2 x+y-3=0$ - This line does not pass through B and the reflection of A.
- Option C: $x-6 y+5=0$ - This line does not pass through B and the reflection of A.
- Option D: none of these - Option B is the correct answer.

Key Concepts:
- Reflection of a point across a line
- Equation of a straight line

Memory Aids:
- Remember that the reflection of a point across y=x swaps the coordinates.

Quick Tips:
- Find the reflection of A in the line y=x and then find the equation of the line passing through that point and B.

175 Easy

175 of 530

Application +4 | -1

If the line

p{x^2} - qxy - {y^2} = 0

makes an angles

\alpha

and

\beta

with x-axis, then the value of tan

\left( {\alpha + \beta } \right)

is

View answer and explanation

Correct option: A

\frac{{ - q}}{{1 + p}}

The correct answer is A ((-q)/(1+p)) because the slopes of the lines are used to find the tangent of the sum of the angles.

Solution approach

Correct Answer + Core Reason: $\frac{{ - q}}{{1 + p}}$ . The slopes of the lines are used.

Step-by-Step Solution:

Step 1: The equation is $px^2 - qxy - y^2 = 0 \Rightarrow y^2 + qxy - px^2 = 0$ .

Step 2: Let the slopes be $m_1$ and $m_2$ . Then $m_1 + m_2 = -q$ and $m_1m_2 = -p$ .

Step 3: $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{m_1 + m_2}{1 - m_1m_2}$ .

Step 4: Substitute the values: $\tan(\alpha + \beta) = \frac{-q}{1 - (-p)} = \frac{-q}{1 + p}$ .

Why other options are wrong:
- Option B: Incorrect sign.
- Option C: Incorrect variables.
- Option D: Incorrect variables and sign.

Key Concepts:
- Slopes of lines represented by a homogeneous equation.
- Tangent of the sum of angles.

Memory Aids:
- Remember that $m_1 + m_2 = -\frac{\text{coefficient of xy}}{\text{coefficient of }y^2}$ and $m_1m_2 = \frac{\text{coefficient of }x^2}{\text{coefficient of }y^2}$ .

Quick Tips:
- Be careful with the signs when using the slope formulas.

176 Easy

176 of 530

Application +4 | -1

Equation of the line passing through the point (

a-1

,

a+1)

and making zero intercept on both axes is

View answer and explanation

Correct option: D

(a+1) x-(a-1) y=0

Option D (

(a+1) x-(a-1) y=0

) is correct because the line passes through the origin and the given point, and this equation satisfies both conditions.

Solution approach

Correct Answer + Core Reason: $(a+1) x-(a-1) y=0$ . The line passes through the origin and the given point.

Step-by-Step Solution:

Step 1: Since the line makes zero intercept on both axes, it passes through the origin (0,0).

Step 2: The line also passes through the point (a-1, a+1).

Step 3: The slope of the line passing through (0,0) and (a-1, a+1) is: $m = \frac{{(a+1) - 0}}{{(a-1) - 0}} = \frac{{a+1}}{{a-1}}$

Step 4: The equation of the line is given by $y = mx$ , so $y = \frac{{a+1}}{{a-1}}x$

Step 5: Rearrange the equation to get: $(a-1)y = (a+1)x$ , which simplifies to $(a+1)x - (a-1)y = 0$

Why other options are wrong:
- Option A: $a x+a y-1=0$ does not pass through the origin (0,0).
- Option B: $(a+1) x+(a-1) y=0$ does not satisfy the given point (a-1, a+1).
- Option C: $(a-1) x-(a+1) y=0$ does not satisfy the given point (a-1, a+1).

Key Concepts:
- Equation of a line
- Slope of a line
- Intercepts

Memory Aids:
- A line with zero intercepts passes through the origin.

Quick Tips:
- If a line passes through the origin, its equation is of the form y = mx.

177 Medium

177 of 530

Application +4 | -1

Let

{ax}+{by}+{c}=0

be a variable straight line, where

{a}, {b}

and

{c}

are

1^{\text {st}}, 3^{\text {rd}}

and

7^{\text {th}}

terms of an increasing A.P. respectively. Then the variable straight line always passes through a fixed point which lies on-

View answer and explanation

Correct option: D

x^2+y^2=13

Option D (

x^2+y^2=13

) is correct because the variable line always passes through a fixed point (2,-3), which lies on the circle

x^2+y^2=13

.

Solution approach

Correct Answer + Core Reason: $x^2+y^2=13$ . The fixed point through which the line passes satisfies this equation.

Step-by-Step Solution:

Step 1: Express b and c in terms of a and d. Given that a, b, and c are the 1st, 3rd, and 7th terms of an increasing arithmetic progression (A.P.), we can write $b = a + 2d$ and $c = a + 6d$ , where d is the common difference.

Step 2: Substitute b and c into the equation of the line. The equation of the line is $ax + by + c = 0$ . Substituting the expressions for b and c, we get $ax + (a + 2d)y + (a + 6d) = 0$ .

Step 3: Rearrange the equation. Rearranging the equation, we have $ax + ay + 2dy + a + 6d = 0$ , which can be written as $a(x + y + 1) + d(2y + 6) = 0$ .

Step 4: Find the fixed point. For the line to always pass through a fixed point, the coefficients of a and d must be zero simultaneously. So, we have the following system of equations:
$x + y + 1 = 0$ and $2y + 6 = 0$ .\nSolving for y, we get $2y = -6$ , so $y = -3$ . Substituting this into the first equation, we get $x - 3 + 1 = 0$ , so $x = 2$ . Thus, the fixed point is $(2, -3)$ .

Step 5: Check which equation the fixed point satisfies. Now, we need to check which of the given options the point $(2, -3)$ satisfies:
- A. $y^2 = 4x$ : $(-3)^2 = 9$ and $4(2) = 8$ . This is not true.
- B. $x^2 + y^2 = 5$ : $(2)^2 + (-3)^2 = 4 + 9 = 13$ . This is not equal to 5.
- C. $3x + 4y = 9$ : $3(2) + 4(-3) = 6 - 12 = -6$ . This is not equal to 9.
- D. $x^2 + y^2 = 13$ : $(2)^2 + (-3)^2 = 4 + 9 = 13$ . This is true.

Final Answer: The fixed point $(2, -3)$ lies on the circle $x^2 + y^2 = 13$ .

Why other options are wrong:
- Option A: $y^2=4 x$ The point (2,-3) does not satisfy this equation.
- Option B: $x^2+y^2=5$ The point (2,-3) does not satisfy this equation.
- Option C: $3 x+4 y=9$ The point (2,-3) does not satisfy this equation.

Key Concepts:
- Arithmetic progression
- Equation of a straight line
- Finding a fixed point

Memory Aids:
- Remember the general form of an A.P.: $a, a + d, a + 2d, ...$

Quick Tips:
- When dealing with a family of lines, try to find a fixed point by setting the coefficients of the parameters to zero.

178 Hard

178 of 530

Application +4 | -1

The algebraic sum of the perpendicular distances from

A\left(x_1, y_1\right), B\left(x_2, y_2\right)

and

C\left(x_3, y_3\right)

to a variable line is zero, then the line passes through

View answer and explanation

Correct option: B

the centroid of

\triangle A B C

Option B is correct. If the algebraic sum of the perpendicular distances from

A(x_1, y_1)

,

B(x_2, y_2)

, and

C(x_3, y_3)

to a variable line is zero, then the line passes through the centroid of

\triangle ABC

.

Solution approach

Correct Answer + Core Reason: the centroid of $\triangle A B C$ . The line passes through the centroid because the sum of perpendicular distances is zero.

Step-by-Step Solution:

Step 1: Let the equation of the variable line be $Px + Qy = 1$ .

Step 2: The perpendicular distance from a point $(x_i, y_i)$ to the line is $\frac{Px_i + Qy_i - 1}{\sqrt{P^2 + Q^2}}$ .

Step 3: The algebraic sum of the perpendicular distances from $A(x_1, y_1)$ , $B(x_2, y_2)$ , and $C(x_3, y_3)$ is zero:
$\frac{Px_1 + Qy_1 - 1}{\sqrt{P^2 + Q^2}} + \frac{Px_2 + Qy_2 - 1}{\sqrt{P^2 + Q^2}} + \frac{Px_3 + Qy_3 - 1}{\sqrt{P^2 + Q^2}} = 0$

Step 4: Simplify the equation:
$P(x_1 + x_2 + x_3) + Q(y_1 + y_2 + y_3) - 3 = 0$
$P(x_1 + x_2 + x_3) + Q(y_1 + y_2 + y_3) = 3$

Step 5: Divide by 3:\nP( $\frac{x_1 + x_2 + x_3}{3}$ ) + Q( $\frac{y_1 + y_2 + y_3}{3}$ ) = 1 $\n\n**Step 6:** This shows that the point$ ( $\frac{x_1 + x_2 + x_3}{3}$ , $\frac{y_1 + y_2 + y_3}{3}$ ) $lies on the line$ Px + Qy = 1 $.\n\n**Step 7:** The point$ ( $\frac{x_1 + x_2 + x_3}{3}$ , $\frac{y_1 + y_2 + y_3}{3}$ ) $is the centroid of the triangle$ ABC $.\n\n**Why other options are wrong:**\n- **Option A:** Incorrect. The orthocenter does not satisfy the given condition.\n- **Option C:** Incorrect. The circumcenter does not satisfy the given condition.\n- **Option D:** Incorrect because one of the other options is correct.\n\n**Key Concepts:**\n- Perpendicular distance from a point to a line\n- Centroid of a triangle\n\n**Memory Aids:**\n- Remember the formula for the centroid of a triangle.\n\n**Quick Tips:**\n- Use the formula for the perpendicular distance from a point to a line.\n- Recognize the coordinates of the centroid.$

179 Easy

179 of 530

Application +4 | -1

The locus of the point of intersection of lines

x\cos \alpha + y\sin \alpha = a

and

x\sin \alpha - y\cos \alpha = b

is (

\alpha

is a variable)

View answer and explanation

Correct option: C

{x^2} + {y^2} = {a^2} + {b^2}

Option C (

{x^2} + {y^2} = {a^2} + {b^2}

) is correct because it represents the locus of the point of intersection of the given lines after eliminating the variable

\alpha

.

Solution approach

Correct Answer + Core Reason: ${x^2} + {y^2} = {a^2} + {b^2}$ . This equation represents the locus of the intersection point of the given lines.

Step-by-Step Solution:

Step 1: Let (h, k) be the point of intersection of the lines $x\cos \alpha + y\sin \alpha = a$ and $x\sin \alpha - y\cos \alpha = b$ .

Step 2: Then, $h\cos \alpha + k\sin \alpha = a$ ... (i) and $h\sin \alpha - k\cos \alpha = b$ ... (ii).

Step 3: Square both equations: $(h\cos \alpha + k\sin \alpha)^2 = a^2$ and $(h\sin \alpha - k\cos \alpha)^2 = b^2$ .

Step 4: Add the squared equations: $(h\cos \alpha + k\sin \alpha)^2 + (h\sin \alpha - k\cos \alpha)^2 = a^2 + b^2$ .

Step 5: Expand and simplify: $h^2\cos^2 \alpha + 2hk\cos \alpha \sin \alpha + k^2\sin^2 \alpha + h^2\sin^2 \alpha - 2hk\sin \alpha \cos \alpha + k^2\cos^2 \alpha = a^2 + b^2$ .

Step 6: Combine terms: $h^2(\cos^2 \alpha + \sin^2 \alpha) + k^2(\sin^2 \alpha + \cos^2 \alpha) = a^2 + b^2$ .

Step 7: Since $\cos^2 \alpha + \sin^2 \alpha = 1$ , we have $h^2 + k^2 = a^2 + b^2$ .

Step 8: Replace (h, k) with (x, y) to find the locus: $x^2 + y^2 = a^2 + b^2$ .

Step 9: Final Answer: ${x^2} + {y^2} = {a^2} + {b^2}$

Why other options are wrong:
- **Option A ( $2({x^2} + {y^2}) = {a^2} + {b^2}$ ):** This equation is incorrect because the coefficients of $x^2$ and $y^2$ should be 1 after simplification.
- **Option B ( ${x^2} - {y^2} = {a^2} - {b^2}$ ): This equation is incorrect because it involves subtraction of squares, which does not result from the given equations.
- Option D (None of these): This is incorrect because option C is the correct answer.

Key Concepts:**
- Trigonometric identities: $\sin^2 \theta + \cos^2 \theta = 1$ .
- Locus of a point.
- Elimination of a variable.

Memory Aids:
- Remember the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ .

Quick Tips:
- Square and add the equations to eliminate trigonometric functions.
- Replace (h, k) with (x, y) to obtain the locus.

180 Easy

180 of 530

Analyzing +4 | -1

If

t_1, t_2

and

t_3

are distinct, the points

\left(t_1, 2 a t_1+a t_1^3\right)

,

\left(t_2, 2 a t_2+a t_2^3\right)

and

\left(t_3, 2 a t_3+a t_3^3\right)

are collinear if

View answer and explanation

Correct option: C

t_1+t_2+t_3=0

Option C is correct. The points are collinear if the determinant formed by their coordinates is zero, which simplifies to t1 + t2 + t3 = 0.

Solution approach

Correct Answer + Core Reason: $t_1+t_2+t_3=0$ . The collinearity condition leads to this relationship between $t_1, t_2,$ and $t_3$ .

Step-by-Step Solution:

Step 1: The points $(t_1, 2at_1 + at_1^3)$ , $(t_2, 2at_2 + at_2^3)$ , and $(t_3, 2at_3 + at_3^3)$ are collinear if the determinant of the matrix formed by their coordinates and 1 is zero:
$\begin{vmatrix} t_1 & 2at_1 + at_1^3 & 1 \\ t_2 & 2at_2 + at_2^3 & 1 \\ t_3 & 2at_3 + at_3^3 & 1 \end{vmatrix} = 0$

Step 2: Factor out $a$ from the second column:
$a \begin{vmatrix} t_1 & 2t_1 + t_1^3 & 1 \\ t_2 & 2t_2 + t_2^3 & 1 \\ t_3 & 2t_3 + t_3^3 & 1 \end{vmatrix} = 0$

Step 3: Apply the row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$ :
$a \begin{vmatrix} t_1 & 2t_1 + t_1^3 & 1 \\ t_2 - t_1 & 2(t_2 - t_1) + (t_2^3 - t_1^3) & 0 \\ t_3 - t_1 & 2(t_3 - t_1) + (t_3^3 - t_1^3) & 0 \end{vmatrix} = 0$

Step 4: Factor out $(t_2 - t_1)$ and $(t_3 - t_1)$ from the second and third rows, respectively:
$a(t_2 - t_1)(t_3 - t_1) \begin{vmatrix} t_1 & 2t_1 + t_1^3 & 1 \\ 1 & 2 + t_2^2 + t_1^2 + t_1t_2 & 0 \\ 1 & 2 + t_3^2 + t_1^2 + t_1t_3 & 0 \end{vmatrix} = 0$

Step 5: Expand the determinant along the third column:
$a(t_2 - t_1)(t_3 - t_1)[(2 + t_3^2 + t_1^2 + t_1t_3) - (2 + t_2^2 + t_1^2 + t_1t_2)]$

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