Straight Line and Pair of Straight Lines 1 Question 60
60. Let $A B C$ be a triangle with $A B=A C$. If $D$ is mid point of $B C$, the foot of the perpendicular drawn from $D$ to $A C$ and $F$ the mid-point of $D E$. Prove that $A F$ is perpendicular to $B E$.
$(1989,5 M)$
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Solution:
- Let $B C$ be taken as $X$-axis with origin at $D$, the mid-point of $B C$ and $D A$ will be $Y$-axis.
Given, $A B=A C$
Let $B C=2 a$, then the coordinates of $B$ and $C$ are $(-a, 0)$ and $(a, 0)$ let $A(0, h)$.
Then, equation of $A C$ is
$$ \frac{x}{a}+\frac{y}{h}=1 $$
and equation of $D E \perp A C$ and passing through origin is
$$ \Rightarrow \quad \begin{aligned} \frac{x}{h}-\frac{y}{a} & =0 \\ x & =\frac{h y}{a} \end{aligned} $$
On solving, Eqs. (i) and (ii), we get the coordinates of point $E$ as follows
$$ \frac{h y}{a^{2}}+\frac{y}{h}=1 \Rightarrow y=\frac{a^{2} h}{a^{2}+h^{2}} $$
$\therefore$ Coordinate of $E=\frac{a h^{2}}{a^{2}+h^{2}}, \frac{a^{2} h}{a^{2}+h^{2}}$
Since, $F$ is mid-point of $D E$.
$\therefore$ Coordinate of $F \frac{a h^{2}}{2\left(a^{2}+h^{2}\right)}, \frac{a^{2} h}{2\left(a^{2}+h^{2}\right)}$
$\therefore$ Slope of $A F$,
$$ \begin{aligned} m _1= & \frac{h-\frac{a^{2} h}{2\left(a^{2}+h^{2}\right)}}{0-\frac{a h^{2}}{2\left(a^{2}+h^{2}\right)}}=\frac{2 h\left(a^{2}+h^{2}\right)-a^{2} h}{-a h^{2}} \\ \Rightarrow \quad m _1 & =\frac{-\left(a^{2}+2 h^{2}\right)}{a h} \end{aligned} $$
and slope of $B E, m _2=\frac{\frac{a^{2} h}{a^{2}+h^{2}}-0}{\frac{a h^{2}}{a^{2}+h^{2}}+a}=\frac{a^{2} h}{a h^{2}+a^{3}+a h^{2}}$
$$ \Rightarrow \quad m _2=\frac{a h}{a^{2}+2 h^{2}} $$
From Eqs. (iii) and (iv), $m _1 m _2=-1 \Rightarrow A F \perp B E$
