Question

If a straight line move so that the sum of the perpendiculars let fall on it from the two fixed points $(3,4)$ and $(7,2)$ is equal to three times the perpendicular on it from a third fixed point $(1,3)$, prove that there is another fixed point through which this line always passes and find its coordinates.

Answer 1

Let the straight line be $ax+by+c=\mathrm{0..........}\left(i\right)$

Perpendicular distance from $(3,4)$ is

$p_1=\frac{3a+4b+c}{\sqrt{a^2+b^2}}$

Perpendicular distance from $(7,2)$ is

$p_2=\frac{7a+2b+c}{\sqrt{a^2+b^2}}$

Perpendicular distance from $(1,3)$ is

$p_3=\frac{a+3b+c}{\sqrt{a^2+b^2}}$

Given $p_1+p_2=3p_3$

$\frac{3a+4b+c}{\sqrt{a^2+b^2}}+\frac{7a+2b+c}{\sqrt{a^2+b^2}}=3\frac{a+3b+c}{\sqrt{a^2+b^2}}3a+4b+c+7a+2b+c=3(a+3b+c)10a+6b+2c=3a+9b+3c7a-3b-c=0-7a+3b+c=0$

Comparing with $\left(i\right)$

$\Rightarrow x=-7,y=3$

So it passes through $(-7,3)$