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

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

Class 11 Mathematics Straight Line Problems related to triangle and quadrilateral

Medium

If the coordinates of 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
$2x + y = 0$
B
$x + 2y = 0$
C
$x - 2y = 0$
D
$2x - y = 0$

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Problems related to triangle and quadrilateral

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Let $ABCD$ be a square of side length $1$. Let $P, Q, R, S$ be points in the interiors of the sides $AD, BC, AB, CD$ respectively,such that $PQ$ and $RS$ intersect at right angles. If $PQ = \frac{3\sqrt{3}}{4}$,then $RS$ equals

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Let $PQR$ be a triangle in which $PQ=3$. From the vertex $R$,draw the altitude $RS$ to meet $PQ$ at $S$. Assume that $RS=\sqrt{3}$ and $PS=QR$. Then,$PR$ equals

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If the coordinates of 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:

Similar Questions

If the sides of a triangle are given by the equations $x + y - 5 = 0$,$x - y + 1 = 0$,and $y - 1 = 0$,then what is its circumcenter?

The coordinates of the feet of the perpendiculars from the vertices of a triangle to the opposite sides are $(20, 25)$,$(8, 16)$,and $(8, 9)$. Find the orthocenter of the triangle.

Let $ABCD$ be a trapezium with $AD$ parallel to $BC$. Assume there is a point $M$ in the interior of the segment $BC$ such that $AB=AM$ and $DC=DM$. Then,the ratio of the area of the trapezium to the area of $\triangle AMD$ is

Let $ABCD$ be a square of side length $1$. Let $P, Q, R, S$ be points in the interiors of the sides $AD, BC, AB, CD$ respectively,such that $PQ$ and $RS$ intersect at right angles. If $PQ = \frac{3\sqrt{3}}{4}$,then $RS$ equals

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