Find the equations of two straight lines drawn through the point $(0,a)$ on which the perpendiculars drawn from the point $(2a,2a)$ are each of length $a$.
My Attempt: The equation of line passing through $(0,a)$ is $y=mx+a$ where $m$ is the slope. How do I get the value of $m$?
On
Solution,
Given,
Points: (0,a) and (2a, 2a)
Now,
y - y1 = m (x - x1)
or, y - a = m (x - 0)
or, mx - y + a = 0.........(i)
Since, perpendicular distance of (i) from (2a, 2a) is a.
a = +- 2am - 2a + a / (root under) m ^ 2 + (-1) ^ 2
or, a (root under) m ^ 2 + (-1) ^ 2 = +- a (2m - 1)
or, (root under) m ^ 2 + (-1) ^ 2 = +- (2m - 1)
Squaring on both sides,
m ^ 2 + 1 = (2m - 1) ^ 2
or, m ^ 2 + 1 = 4m ^ 2 - 4m + 1
or, m ^ 2 = 4m ^ 2 - 4m
or, 4m ^ 2 - m ^ 2 = 4m
or, 3m ^ 2 = 4m
or, 3m ^ 2 - 4m = 0
or, m (3m - 4) = 0
Either, m = 0 Or, 3m - 4 = 0
or, 3m = 4
or, m = 4 / 3
Putting m = 0 in equation (i),
0.x - y + a = 0
or, y = a
Putting m = 4 / 3 in equation (i),
4 / 3x - y + a = 0
or, 4x - 3y + 3a / 3 = 0
So, 3y = 4x + 3a
Therefore, the required equations are y = a and 3y = 4x + 3a.
Well, recall the length of a perpendicular from point $(x_{1},y_{1})$ to the line $ax+by+c=0$ is equal to the distance between them and is thus given by $$\dfrac{|ax_{1}+by_{1}+c|}{\sqrt{a^2+b^2}}$$
So the length of a perpendicular from point $(2a,2a)$ on the line $mx-y+a=0$ would be given by $$\dfrac{|2ma-a|}{\sqrt{m^2+1}}=a$$ So squaring and dividing by $a^2$ we have $$\dfrac{4m^2-4m+1}{m^2+1}=1 \implies m=0 \text{ or } m=\dfrac{4}{3}$$