Question

From a point $P=(3,4)$ perpendiculars $PQ$ and $PR$ are drawn to line $3x+4y-7=0$ and a variable line $y-1=m(x-7)$ respectively, then maximum area of $△PQR$ is

• A. $10$
• B. $12$
• C. $6$
• D. $9$

Answer 1

Answer: (D)

$9$

Given

Point $P(3,4)$ and PQ is perpendicular on line $3x+4y-7-0$

SO perpendicular distance

$PQ=\left|\frac{3\left(3\right)+4\left(4\right)-7}{\sqrt{3^2+4^2}}\right|$

$PQ=\left|\frac{9+16-7}{\sqrt{25}}\right|$

$PQ=\frac{18}{5}$

PR is perpendicular from line $y-1=m(x-7)$

$PR=\left|\frac{m\left(3\right)-4-7m+1}{\sqrt{m^2+(-1{)}^2}}\right|$

$PR=\left|\frac{-4m-3}{\sqrt{m^2+1}}\right|$

$PR=\frac{4m+3}{\sqrt{m^2+1}}$

$area=\frac{1}{2}\times PQ\times PR$

$area=\frac{1}{2}\times \frac{18}{5}\times \frac{4m+3}{\sqrt{m^2+1}}$

maximum value of area depends on maximum value of $\frac{4m+3}{\sqrt{m^2+1}}$

Here the slope of $PQ=\tan \theta =\frac{4}{5}$

So in traingle by hypotenious theoram

maximum distance $PR=5$

$area=\frac{1}{2}\times \frac{18}{5}\times 5$

$area=9$