Straight lines practice · Mixed difficulty

Option B (${p_1} = {p_2}$) is correct because the line $2x + 11y = 5$ is the angle bisector of the two lines, hence the perpendicular distances are equal.

Option C ($(5 / 6,6)$) is correct because it satisfies the equation of the perpendicular bisector of the line segment QR, which is the locus of point P.

Option C ($x + 2y = 5$) is correct because the line that maximizes the distance between two points passes through both points.

Option D ($\tan ^{-1} \frac{3}{4}$) is correct because it is calculated using the formula for the angle between the two straight lines represented by the given equation.

Option B (($x^2-y^2$)$^2=16xy$) is correct because it represents the locus of the point derived from the given parametric coordinates.

Option B (${x^2} + {y^2} - 4\left| y \right| = 0$) is correct because it directly represents the locus of a point where 4 times its distance from the x-axis equals the square of its distance from the origin.

The correct answer is D (None of these) because the calculated third vertex does not match any of the given options after applying the orthocenter and centroid conditions.

Option B ($\sqrt {2/3}$) is correct because it is the side length of the equilateral triangle calculated using the perpendicular distance from the vertex to the base.

Option A ($2 = \sqrt 3 r\cos \mathrm{\theta } - 2r\sin \mathrm{\theta }$) is correct because it represents the line perpendicular to the given line and passing through the specified point in polar coordinates.

Option D ((0,0)) is correct because the image of P(a, b) on y = -x is Q(-b, -a), and the image of Q on y = x is R(-a, -b). The midpoint of PR is then ((a - a)/2, (b - b)/2) = (0, 0).

Option B ($\frac{a}{h}+\frac{b}{k}=1$) is correct because it is derived from the condition that the determinant of the matrix formed by the points is zero, indicating collinearity.

Option A ((3,1) and (-7,11)) is correct because these are the points on the line x+y=4 that are at a unit distance from the line 4x+3y-10=0.

Option B ($x\cos \mathrm{\theta } - \;y\sin \mathrm{\theta } = - a\cos 2\mathrm{\theta }$) is correct because it represents the equation of a line passing through the given point and perpendicular to the given line.

The correct answer is C (x + y - 6 = 0) because the equation of a line making equal intercepts on the axes and passing through (2, 4) is found.

Option B ((-a, 0)) is correct because the orthocenter of the triangle formed by the given points lies on the line x = -a, which means the orthocenter is (-a, 0).

Option B ($119x + 102y = 125$) is correct because it represents a line passing through the intersection of the given lines and is perpendicular to the specified line.

Option C ($13\left( {{x^2} + {y^2}} \right) - 83x + 64y + 182 = 0$) is correct because it represents the locus of the point satisfying the given condition.

Option D (Lie on a straight line) is correct because the ratio of the y-coordinates to the x-coordinates is constant, indicating collinearity.

Option C (A circle) is correct because the locus of the midpoint of AB, where A and B move on the coordinate axes such that the distance between them remains constant, is a circle.

Option D ($x - 7y + 6 = 0$) is correct because it represents the other line when 2x-y=1 bisects the angle between it and y=x.

Option B ((-3,2)) is correct because it represents the image of the translated point Q in the line x+y=1.

Option A (Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation of Statement-1) is correct because the angles are equal when the lines are parallel, and this parallelism is implied when the constants are proportional.

Option B ($x \cos \theta-y \sin \theta=a \cos 2 \theta$) is correct because it represents the equation of the line passing through the given point and perpendicular to the given line.

Option D ((4,3),(7,-2)) is correct because it satisfies the median equations and the coordinates of vertex A.

Option D ((1,2)) is correct because the intersection of two perpendicular sides gives the orthocenter.

The correct answer is A (straight line) because after squaring and simplifying, the given equation reduces to a linear equation.

Option B (1) is correct because solving the collinearity condition using the determinant results in a single valid value for $\theta$ within the given interval.

Option A (3) is correct because it is the slope of the line passing through the origin and the midpoint of the line segment joining P and Q.

Option C ($a\mathrm{\alpha } + b\mathrm{\gamma } = 0$) is correct because it represents the condition where the slopes of the two lines are equal.

Option B (1) is correct because only one line in the family is at the specified distance from the point P.