Straight lines practice · Mixed difficulty

Option C ($2 \sqrt{2}$) is correct because the condition for concurrency leads to a quadratic equation in 'm' which must have real roots, thus giving the least value of |a| as $2 \sqrt{2}$.

Option B ($2 x-3 y+1=0$) is correct because it has the same slope as the given line and passes through the point (1,1).

The correct answer is None of the above because the combined equation of lines AB and AC is $3x^2 - 8xy - 3y^2 - 4x + 22y - 7 = 0$, which is not among the options.

Option D ($\left(\frac{{28}}{{5}}, \frac{{1}}{{5}}\right)$) is correct because it is the midpoint of CD, found by solving the equations of line CD and the perpendicular bisector of AB.

Option A ($\sqrt 2$) is correct because it represents the distance calculated by finding the intersection point of the two lines and using the given angle.

Option A (9) is correct because the centroid's coordinates are (4, 5), and (h+k)/(k-h) = (4+5)/(5-4) = 9.

Option B ({2,3,-5}) is correct because these are the values of k for which the determinant of the coefficient matrix is zero, indicating concurrency, but k=2 makes the first two lines parallel.

The correct answer is A (x > 8/15, y < -8/5) because the point P(x, y) lies on the line y = -3x and is on the opposite side of the line 3x - 4y - 8 = 0 as the point (3, 4).

Option B ($\theta=\tan ^{-1}\left(-\frac{\sqrt{57}}{4}\right)$) is correct because it is the angle calculated using the formula for the angle between two lines represented by a homogeneous equation.

Option C ($2 \sqrt{2}$) is correct because the least value of $|a|$ is found by ensuring the discriminant of the quadratic equation in 'm' is non-negative.

Option A ($\sqrt{5}-2 \sqrt{2}) x+(\sqrt{5}+\sqrt{2}) y-3 \sqrt{5}+2 \sqrt{2}=0$) is correct because it represents the angle bisector that contains the point (1,1) after proper normalization.

Option A ($x+4=0$) is correct because the tangents intersect at right angles, which implies that the directrices of the parabolas intersect at the point of intersection of the tangents. The directrices are x = -4 and the intersection point lies on the line x = -4 or x + 4 = 0.

Option B ($3x - 4y + 1 = 0$) is correct because it represents the line in the system farthest from the point (4, -3), which is perpendicular to the line joining (1,1) and (4,-3).

Option A ($\sec ^{-1} p$) is correct because it is calculated using the formula for the angle between the lines and trigonometric identities.

Option A Below the $X$-axis at a distance of $\frac{3}{2}$ from it is correct because the line parallel to the X-axis is found to be y = -3/2.

Option C (obtuse angled) is correct because the cosine of one of the angles is negative, indicating an angle greater than 90 degrees.

Option C (lie on a straight line) is correct because the slope between any two pairs of points is the same, indicating collinearity.

Option C (4) is correct because the product of the slopes of two perpendicular lines is -1, which leads to the equation (1-x)/(3-2) * (5-4)/(7-x) = -1, solving for x gives x=4.

Option B ($\sqrt {10}$) is correct because the radius of the circumcircle is calculated using the distance between the circumcenter and a point on the circumcircle, which is the reflection of the orthocenter in side BC.

Option A (a straight line) is correct because the locus simplifies to a linear equation.

Option A ($\left( {4,\;8/3} \right)$) is correct because it represents the intersection of the altitudes of the triangle.

Option B ($\mathrm{\pi }/3$) is correct because the acute angle is calculated by finding the intersection points and their respective angles with the origin.

Option A $\left(0, \frac{\pi}{2}\right)$ is correct because the values of $\theta$ in this interval ensure that both points lie on the same side of the line x+y-1=0.

Option B ($5x - 6y + 4 = 0$) is correct because it represents the equation of the straight line passing through the intersection points of the two families of lines.

Option A (Lie on a straight line) is correct because the area of the triangle formed by the points is zero, indicating collinearity.

The correct answer is C (8x + 12y - 1 = 0) because it represents a line equidistant from the two given parallel lines, found by averaging their constant terms.

Option D ($x+y=3-2 \sqrt{6}$) is correct because it is the equation of the line passing through the translated point Q in the third quadrant and perpendicular to the given line L.

Option B ((-9,-6)) is correct because the point lies on the line formed by the centroid and the intersection point of the given lines.

Option B ((2,-1)) is correct because the centroid coordinates are found by averaging the x and y coordinates of the vertices, and using Vieta's formulas.

Option A $y=2 x-8$ is correct because it is the equation of the line with slope 2 that passes through the fixed point of the given family of lines.