(C) Given points are $A(-1, 5)$,$B(0, 0)$,and $C(2, 2)$.$D$ is the midpoint of $BC$,so $D = (\frac{0+2}{2}, \frac{0+2}{2}) = (1, 1)$.The slope of line

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(C) Given points are $A(-1, 5)$,$B(0, 0)$,and $C(2, 2)$.$D$ is the midpoint of $BC$,so $D = (\frac{0+2}{2}, \frac{0+2}{2}) = (1, 1)$.The slope of line $AD$ is $m_{AD} = \frac{1-5}{1-(-1)} = \frac{-4}{2} = -2$.The equation of line $AD$ is $y - 5 = -2(x + 1)$,which simplifies to $2x + y - 3 = 0$.The line perpendicular to $AD$ has a slope $m = -\frac{1}{m_{AD}} = \frac{1}{2}$.Since this line passes through $B(0, 0)$,its equation is $y - 0 = \frac{1}{2}(x - 0)$,which simplifies to $x - 2y = 0$.

Class 11 Mathematics Straight Line Foot of perpendicular, Image of a point and Reflexive properties

Medium

If the coordinates of the points $A, B, C$ are $(-1, 5), (0, 0)$ and $(2, 2)$ respectively and $D$ is the midpoint of $BC$,then the equation of the perpendicular drawn from $B$ to the line $AD$ is

A
$x + 2y = 0$
B
$2x + y = 0$
C
$x - 2y = 0$
D
$2x - y = 0$

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Straight Line

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Foot of perpendicular, Image of a point and Reflexive properties

Let $0 < \alpha < \pi/2$ be a constant angle. If $P \equiv (\cos \theta, \sin \theta)$ and $Q \equiv (\cos(\alpha - \theta), \sin(\alpha - \theta))$,how is $Q$ obtained from $P$?

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$A$ ray of light along $x + \sqrt{3}y = \sqrt{3}$ gets reflected upon reaching the $x$-axis. The equation of the reflected ray is:

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The area (in square units) of the quadrilateral formed by the point of intersection of the lines $x+y-1=0$ and $x-y+1=0$,the point $P(1,1)$,and the feet of the perpendiculars from this point onto the lines is:

EasyTS EAMCET 2020

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If the perpendicular drawn from the point $(2, -3)$ to the straight line $4x - 3y + 8 = 0$ meets it at $M(a, b)$ and $a^3 - b^3 = k^3$,then $k=$

MediumTS EAMCET 2025

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$N(3, -4)$ is the foot of the perpendicular drawn from the origin to a line $L$. Then the equation of the line $L$ is

EasyMHT CET 2022

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If the coordinates of the points $A, B, C$ are $(-1, 5), (0, 0)$ and $(2, 2)$ respectively and $D$ is the midpoint of $BC$,then the equation of the perpendicular drawn from $B$ to the line $AD$ is

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Let $0 < \alpha < \pi/2$ be a constant angle. If $P \equiv (\cos \theta, \sin \theta)$ and $Q \equiv (\cos(\alpha - \theta), \sin(\alpha - \theta))$,how is $Q$ obtained from $P$?

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