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JEE • MATHEMATICS • STRAIGHT LINES

Straight lines practice · Mixed difficulty

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Showing 271-300 of 555 questions

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271 Easy
271 of 555
Application +4 | -1
The coordinate of the foot of perpendicular from (a,0) on the line y=mx+amy = mx + \frac{a}{m} are
View answer and explanation
Correct option: A
(0,am)\left( {0,\frac{a}{m}} \right)
Option A (0,am)\left( {0,\frac{a}{m}} \right) is correct because it represents the foot of the perpendicular from (a,0) to the given line, found by solving the equations of the two lines.
272 Easy
272 of 555
Application +4 | -1
Area of polygon having sides y−x−1=0,2y−2x+2=0,y−2x−2=0y-x-1=0,2 y-2 x+2=0, y-2 x-2=0 and 3y−6x−9=03 y-6 x-9=0
View answer and explanation
Correct option: D
2
Option D, 2, is the correct area of the parallelogram formed by the given lines, calculated using the formula for the area of a parallelogram formed by two pairs of parallel lines.
273 Easy
273 of 555
Application +4 | -1
Distance between lines represented by x2+22xy+2y2+4x+42y+1=0x^2+2 \sqrt{2} x y+2 y^2+4 x+4 \sqrt{2} y+1=0
View answer and explanation
Correct option: B
2
The distance between the lines is 2 because the given equation represents a pair of parallel lines, and the distance is calculated using the formula for distance between parallel lines.
274 Easy
274 of 555
Application +4 | -1
Angles made with the xx-axis by two lines drawn through the point (1,2)(1,2) and cutting the line x+y=4x+y=4 at a distance 63\frac{\sqrt{6}}{3} from the point (1,2)(1,2) are
View answer and explanation
Correct option: C
π/12\pi / 12 and 5π/125 \pi / 12
Option C (π/12\pi / 12 and 5π/125 \pi / 12) is correct because these angles satisfy the condition that the lines cut x+y=4 at a distance of 63\frac{\sqrt{6}}{3} from (1,2).
275 Easy
275 of 555
Application +4 | -1
Two points (a,0) and (0,b) are joined by a straight line. Another point on this line, is
View answer and explanation
Correct option: A
(3a,−2b)\left( {3a, - 2b} \right)
Option A ((3a,−2b)\left( {3a, - 2b} \right)) is correct because it satisfies the equation of the line joining the points (a, 0) and (0, b).
276 Easy
276 of 555
Analyzing +4 | -1
The points A(2a,  4a),  B(2a,  6a)A\left( {2a,\;4a} \right),\;B\left( {2a,\;6a} \right) and C(2a+3a,  5a),  a>0C\left( {2a + \sqrt 3 a,\;5a} \right),\;a > 0 are the vertices of
View answer and explanation
Correct option: C
An acute angled triangle
Option C (An acute angled triangle) is correct because the triangle is equilateral, and equilateral triangles are acute angled.
277 Easy
277 of 555
Understanding +4 | -1
Distance between P(x1,y1)P\left(x_1, y_1\right) and Q(x2,y2)Q\left(x_2, y_2\right) when PQP Q is parallel to yy-axis is
View answer and explanation
Correct option: D
∣y1−y2∣\left|y_1-y_2\right|
Option D (∣y1−y2∣\left|y_1-y_2\right|) is correct because when PQ is parallel to the y-axis, the distance is the absolute difference of the y-coordinates.
278 Easy
278 of 555
Application +4 | -1
Points (8,2),(−2,−2)(8,2),(-2,-2) and (3,0)(3,0) are the vertices of
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the points are collinear, meaning they lie on the same line and do not form a triangle.
279 Easy
279 of 555
Application +4 | -1
The distance of the point (2,3)(2,3) from the line 2x−3y+9=02 x-3 y+9=0 measured along the line x−y+1=0,x-y+1=0, is
View answer and explanation
Correct option: D
424 \sqrt{2}
Option D (424 \sqrt{2}) is correct because it is the distance calculated along the line x−y+1=0x-y+1=0 from the point (2,3) to the line 2x−3y+9=02x-3y+9=0.
280 Easy
280 of 555
Application +4 | -1
The equation of the line with gradient −3/2- 3/2 which is concurrent with the lines 4  x+3  y−7=04\;x + 3\;y - 7 = 0 and 8  x+5  y−1=08\;x + 5\;y - 1 = 0 is
View answer and explanation
Correct option: A
3  x+2  y−2=03\;x + 2\;y - 2 = 0
The correct answer is A (3x + 2y - 2 = 0) because it represents a line with the given gradient that passes through the intersection of the two given lines.
281 Easy
281 of 555
Understanding +4 | -1
If the two vartiables xx and yy have a perfect correction (direction indirect), then they may be connected by a relation of the type
View answer and explanation
Correct option: C
xa+yb=1\frac{x}{a}+\frac{y}{b}=1
Option C (xa+yb=1\frac{x}{a}+\frac{y}{b}=1) is correct because it represents a linear relationship with a negative slope, indicating an indirect correlation.
282 Easy
282 of 555
Analyzing +4 | -1
If the points (a1,b1),(a2,b2)\left( {{a_1},{b_1}} \right),\left( {{a_2},{b_2}} \right) and (a3,b3)\left( {{a_3},{b_3}} \right) are collinear, then lines aix+biy+1=0{a_i}x + {b_i}y + 1 = 0 for aix+biy+1=0{a_i}x + {b_i}y + 1 = 0 are
View answer and explanation
Correct option: A
Concurrent
Option A (Concurrent) is correct because the collinearity of the points implies that the determinant formed by their coordinates is zero, which is also the condition for concurrency of the given lines.
283 Easy
283 of 555
Application +4 | -1
The distance of the point (-2,3) from the line x-y=5 is
View answer and explanation
Correct option: A
525\sqrt 2
Option A (525\sqrt 2) is correct because the distance is calculated using the point-to-line distance formula.
284 Easy
284 of 555
Application +4 | -1
The straight line y=x−2y=x-2 rotates about a point where it cuts the x-axis and becomes perpendicular to the straight line ax+by+c=0.ax+by+c=0. Then its equation is
View answer and explanation
Correct option: D
ay−bx+2b=0a y-b x+2 b=0
Option D (ay−bx+2b=0a y-b x+2 b=0) is correct because the new line is perpendicular to the given line and passes through the intersection point with the x-axis.
285 Easy
285 of 555
Application +4 | -1
If non-zero numbers a,b,c are in HP, then the straight line xa+yb+1c=0\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0 always passes throught a fixed point.That point is
View answer and explanation
Correct option: B
(1,  −2)\left( {1,\; - 2} \right)
Option B ((1,  −2)\left( {1,\; - 2} \right)) is correct because it satisfies the condition derived from the harmonic progression.

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286 Medium
286 of 555
Application +4 | -1
Family of a lines xsec⁡2θ+ytan⁡2θ−2=0x \sec ^2 \theta+y \tan ^2 \theta-2=0, for different real θ\theta, is
View answer and explanation
Correct option: C
concurrent at (2,−2)(2,-2)
Option C (concurrent at (2,−2)(2,-2)) is correct because all lines in the family pass through this specific point, regardless of the value of θ\theta.
287 Easy
287 of 555
Application +4 | -1
The straight line whose sum of the intercepts on the axes is equal to half to the product of the intercepts, passes through the point whose coordinates are
View answer and explanation
Correct option: B
(2,2)
Option B (2,2) is correct because the given condition leads to the equation of a line that passes through the point (2,2).
288 Easy
288 of 555
Analyzing +4 | -1
If O′(4,  8/3)O'\left( {4,\;8/3} \right) is the orthocentre of the triangle ABCABC the coordinates of whose vertices are O(0,0),A(8,0)O\left( {0,0} \right),A\left( {8,0} \right) and B(4,6),B\left( {4,6} \right), then the coordinates of the orthocentre of Δ  O′AB\Delta \;O'AB are
View answer and explanation
Correct option: A
(0,  0)\left( {0,\;0} \right)
Option A ((0, 0)) is correct because if O' is the orthocenter of triangle ABC, then O, A, B, and O' are such that each point is the orthocenter of the triangle formed by the remaining three points.
289 Easy
289 of 555
Understanding +4 | -1
The point of intersection of the lines represented by equation 2(x+2)2+3(x+2)(y−2)−2(y−2)2=02(x+2)^2+3(x+2)(y-2)-2(y-2)^2=0 is
View answer and explanation
Correct option: C
(−2,2)(-2,2)
Option C ((-2,2)) is correct because substituting X=x+2 and Y=y-2 transforms the equation to 2X²+3XY-2Y²=0, which intersects at X=0 and Y=0, thus x=-2 and y=2.
290 Easy
290 of 555
Application +4 | -1
If the points (1,3) and (5,1) are two opposite vertices of a rectangle and the other two vertices lie on the line y=2x+c,y = 2x + c,, then the value of c is
View answer and explanation
Correct option: B
-4
Option B (-4) is correct because the midpoint of the given vertices lies on the line y = 2x + c, allowing us to solve for c.
291 Easy
291 of 555
Application +4 | -1
The coordinates of the image of the origin O with respect to the line x+y+1=0x + y + 1 = 0 are
View answer and explanation
Correct option: D
(−1,−1)\left( { - 1, - 1} \right)
Option D ((-1, -1)) is correct because it is the image of the origin calculated using the formula for the image of a point with respect to a line.
292 Easy
292 of 555
Analyzing +4 | -1
The foot of the perpendicular drawn from the origin, on the line, 3x+y=λ(λ≠0)3 x+y=\lambda(\lambda \neq 0) is PP. If the line meets xx-axis at AA and yy-axis at BB, then the ratio BP:PAB P: P A is
View answer and explanation
Correct option: D
3:13: 1
Option D (3:13: 1) is correct because it represents the ratio of BP to PA, calculated using the coordinates of the points and the distance formula.
293 Easy
293 of 555
Application +4 | -1
If a,b,ca, b, c are in A.P., then ax+by+c=0a x+b y+c=0 represents
View answer and explanation
Correct option: B
a family of concurrent lines
Option B (a family of concurrent lines) is correct because the given equation can be rewritten to represent lines passing through the intersection of two other lines.
294 Easy
294 of 555
Application +4 | -1
The distance between the parallel lines y=x+a,  y=x+by = x + a,\;y = x + b is
View answer and explanation
Correct option: A
∣b−a∣2\frac{{\left| {b - a} \right|}}{{\sqrt 2 }}
The correct answer is A (|b - a|/√2) because the distance between two parallel lines is calculated using the formula.
295 Medium
295 of 555
Application +4 | -1
Consider a family of straight lines (x+y)+λ(2x−y+1)=0(x+y)+\lambda(2 x-y+1)=0. Equation of the straight line belonging to this family that is farthest from (1,−3)(1,-3) is
View answer and explanation
Correct option: D
15y−6x=715 y-6 x=7
Option D (15y−6x=715y - 6x = 7) is correct because it represents the line in the family that is perpendicular to the line connecting the point of concurrency and the given point.
296 Easy
296 of 555
Application +4 | -1
If the points (−2,  −5),(2,  −2),  (8,  a)\left( { - 2,\; - 5} \right),\left( {2,\; - 2} \right),\;\left( {8,\;a} \right) are collinear, then the value of aa is
View answer and explanation
Correct option: B
52\frac{5}{2}
Option B (52\frac{5}{2}) is correct because the slope between any two points on a line must be equal if the points are collinear.
297 Easy
297 of 555
Application +4 | -1
A straight line through P(1,2) is such that its intercept between the axes is bisected at P. Its equation is
View answer and explanation
Correct option: D
2x+y=42x + y = 4
Option D (2x+y=42x + y = 4) is correct because it represents the equation of the line whose intercept between the axes is bisected at P(1,2).
298 Medium
298 of 555
Application +4 | -1
A line AB makes zero intercepts on x-axis and y-axis and it is perpendicular to another line CD which is 3x+4y+6=03x + 4y + 6 = 0. The equation of line AB is
View answer and explanation
Correct option: C
4x−3y=04x - 3y = 0
Option C (4x−3y=04x - 3y = 0) is correct because the line passes through the origin and is perpendicular to the given line.
299 Medium
299 of 555
Application +4 | -1
Number of lines drawn from the point (4,−5)(4,-5) so that its distance from (−2,3)(-2,3) will be equal to 12 are
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the discriminant of the resulting quadratic equation is negative, indicating no real solutions for the slope 'm', and thus no such lines exist.
300 Easy
300 of 555
Application +4 | -1
Circumcentre of triangle whose vertices are (0, 0), (3, 0) and (0, 4) is
View answer and explanation
Correct option: A
(32,2)\left( {\frac{3}{2},2} \right)
Option A (32,2)\left( {\frac{{3}}{2},2} \right) is correct because the circumcentre of a right-angled triangle is the midpoint of the hypotenuse.
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