Find the equations of the two straight lines drawn through the point $(0, a)$ on which the perpendiculars let fall from the point $(2 a, 2 a)$ are each of length $a$ . Prove also that the equation of the straight line joining the feet of these perpendiculars is $y + 2 x = 5 a$ .

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Solution

Let the equation of straight line be $y = m x + c$

It passes through $(0, a)$

$a = m (0) + c \Rightarrow c = a y = m x + a . . . . . . . . (i) m x - y + a = 0$

Lenght of perpendicular from $(2 a, 2 a)$ is equal to $a$

$\frac{| 2 a m - 2 a + a |}{\sqrt{1 + m^{2}}} = a | 2 a m - a | = a \sqrt{1 + m^{2}} 4 a^{2} m^{2} + a^{2} - 4 a^{2} m = a^{2} + a^{2} m^{2} 3 a^{2} m^{2} - 4 a^{2} m = 0 3 m^{2} - 4 m = 0 m (3 m - 4) = 0 \Rightarrow m = 0, \frac{4}{3}$

Substituting $m$ in $(i)$

$m = 0$

$y = 0 (x) + a y = a$

$m = \frac{4}{3}$

$y = \frac{4}{3} x + a 3 y = 4 x + 3 a . . . . (i i)$

Slope of line perpendicular to $(i)$ is

$m^{'} = - \frac{3}{4}$

Equation of line passing $(2 a, 2 a)$ and slope $m^{'}$ is

$y - 2 a = - \frac{3}{4} (x - 2 a) 4 y - 8 a = - 3 x + 6 a 3 x + 4 y = 14 a . . . . . . (i i i)$

Feet of perpendicular is point of intersection of $(i i)$ and $(i i i)$

Solving $(i i)$ and $(i i i)$ , we get

$P (\frac{6 a}{5}, \frac{13 a}{5})$

Feet of perpendicular from $(2 a, 2 a)$ on $y = a$ is

$Q (2 a, a)$

Equation of $P Q$ is

$y - a = \frac{\frac{13 a}{5} - a}{\frac{6 a}{5} - 2 a} (x - 2 a) y - a = - 2 (x - 2 a) y + 2 x = 5 a$

Hence proved.

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Find the equations of the two straight lines drawn through the point (0,a) on which the perpendiculars let fall from the point (2a,2a) are each of length a. Prove also that the equation of the straight line joining the feet of these perpendiculars is y+2x=5a.

Find the equations of the two straight lines drawn through the point $(0, a)$ on which the perpendiculars let fall from the point $(2 a, 2 a)$ are each of length $a$ . Prove also that the equation of the straight line joining the feet of these perpendiculars is $y + 2 x = 5 a$ .