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The hypotenuse of a right angled triangle has its ends at the points $(1,3)$ and $(-4,1)$. Find the equation of the legs (Perpendicular sides) of the triangle.
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Let $A B C$ be the right angled triangle such that $\angle C=90^{\circ}$
Let $m$ be the slope of $A C$
$\therefore$ the slope of $B C=-\frac{1}{m}$
$\therefore \quad$ Equation of $A C$ is, $y-3=m(x-1)$
$\Rightarrow \quad x-1=\frac{1}{m}(y-3)$
Equation of BC, $x+4=-m(y-1)$
$\Rightarrow y-1=-\frac{1}{m}(x+4)$
For $m=0$ these line are $x+4=0, y-3=0$
For $m=\infty$ the lines are $x-1=0, y-1=0$
Let $m$ be the slope of $A C$
$\therefore$ the slope of $B C=-\frac{1}{m}$
$\therefore \quad$ Equation of $A C$ is, $y-3=m(x-1)$
$\Rightarrow \quad x-1=\frac{1}{m}(y-3)$
Equation of BC, $x+4=-m(y-1)$
$\Rightarrow y-1=-\frac{1}{m}(x+4)$
For $m=0$ these line are $x+4=0, y-3=0$
For $m=\infty$ the lines are $x-1=0, y-1=0$
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