Show that the perpendicular drawn from the point (4,1) on the lin… - Vidyadip

Question

Show that the perpendicular drawn from the point $(4,1)$ on the line segment joining $(6,5)$ and $(2,-1)$ divide it internally in the ratio $8: 5$.

✓

Answer

Let the points be $A(4,1), B(6$, 5) and $C(2,-1)$

$A P \perp B C$ is drawn Slope of line $B C$,
$m=\frac{-1-5}{2-6}=\frac{3}{2}$
$\begin{array}{l}\text { Since } A P \perp B C \\
\therefore \text { Slope of } A P,\end{array}$
$m^{\prime}=-\frac{1}{m}=-\frac{2}{3}$
$\therefore$ Equation of line $A P$ is
$y-1=-\frac{2}{3}(x-4)$
⇒ $2 x+3 y-11=0\ldots(i)$
Let P divide BC in the ratio m : n.
$\therefore$ Coordinates of $P$ are $\left(\frac{2 m+6 n}{m+n}, \frac{-m+5 n}{m+n}\right)$
Since P lies on $A P$, so (i)
⇒ 2. $\frac{2 m+6 n}{m+n}+3 \cdot\left(\frac{-m+5 n}{m+n}\right)-11=0$
$\Rightarrow 4 m+12 n-3 m+15 n-11 m-11 n=0$
⇒ $-10 m+16 n=0$
⇒ $\frac{m}{n}=\frac{16}{10}$
⇒ $m: n=8: 5$.
Hence Proved.

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