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Q.
The hypotenuse of a right angled triangle has its ends at the points $(1,3)$ and $(-4,1)$. The legs (perpendicular sides) of the triangle are
Straight Lines
Solution:
First we plot the points $A(1,3)$ and $B(-4,1)$ in the $x y$-plane. From the point $A(1,3)$, we draw a line parallel to $Y$-axis and from the point $B(-4,1)$, we draw a line parallel to $X$-axis. The point of intersection of two lines is on $C$, which is right angled at $C$.
$\therefore$ The coordinate of $C$ will be $(1,1)$.
$\therefore$ Equation of line $A C$ passing through $A(1,3)$ and $C(1,1)$ is
$ y-y_1 =\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$
$\therefore y-3 =\frac{1-3}{1-1}(x-1) $
$\Rightarrow y-3=\frac{-2}{0}(x-1) \Rightarrow x=1$
Equation of line $B C$ is
$y-1=\frac{1-1}{1+4}(x-1)$
$ \Rightarrow y-1=\frac{0}{1+4}(x-1) $
$\Rightarrow y-1=0 \Rightarrow y=1 $
Hence, the legs of a triangle are $x=1$ and $y=1$.
