Question

Find the image of point $(3,4)$ w.r.t the line $x-4y-4=0$

• A. $(\frac{-21}{17},\frac{-52}{17})$
• B. $(\frac{21}{17},\frac{-52}{17})$
• C. $(\frac{-21}{17},\frac{52}{17})$
• D. $(\frac{21}{17},\frac{52}{17})$

Answer 1

The correct option is D $(\frac{21}{17},\frac{52}{17})$

Let $P(a,b)$ be the image of given point $A(3,4)$ w.r.t the line $x-4y-4=0$

Now,

mid point of $AP=(\frac{a+3}{2},\frac{b+4}{2})$

mid point of $AP$ must be lie on line $x-4y-4=0$

$\Rightarrow \left(\frac{a+3}{2}\right)-4\left(\frac{b+4}{2}\right)-4=0$

$\Rightarrow a+3-4b+16-8=0$

$\Rightarrow a-4b=-11$ .......... eq (i)

Now,

Slope of $AP=\frac{b-4}{a-3}$

Slope of line $=\frac{1}{4}$

$AP$ and the line must be perpendicular to each other

$\Rightarrow$ Slope of $AP\times$ Slope of line $=-1$

$\Rightarrow \frac{b-4}{a-3}\times \frac{1}{4}=-1$

$\therefore 4a+b=8$ ..........eq(ii)

Solving equation (i) and (ii) we get

$a=\frac{-3}{17}$

$b=\frac{46}{17}$

Therefore, $(\frac{-3}{17},\frac{46}{17})$ is the image of point $(3,4)$