A line passes through the origin and makes equal angles with the positive coordinate axes.

Question

A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $L_1: 2x+y+6=0$ and $L_2: 4x+2y-p=0,\ p>0$, at the points $A$ and $B$, respectively. If $AB=\dfrac9{\sqrt2}$ and the foot of the perpendicular from the point $A$ on the line $L_2$ is $M$, then $\dfrac{AM}{BM}$ is equal to:

Accepted Answer

The line $y=x$ makes angle $\theta$ with $L_2$ (slope $-2$) where $\tan\theta = \left|\dfrac{-2-1}{1-2}\right|=3$, giving $\sin\theta=\dfrac3{\sqrt{10}}$.

$$AM = AB\sin\theta = \frac9{\sqrt2}\cdot\frac3{\sqrt{10}} = \frac{27}{2\sqrt5}$$

Since $L_1\parallel L_2$, using the perpendicular distance relation, solving for $p$ (with $p>0$) gives $p=15$, and then
$$BM = \sqrt{\left(\frac9{\sqrt2}\right)^2-\left(\frac{27}{2\sqrt5}\right)^2} = \frac{9}{2\sqrt5}$$

$$\frac{AM}{BM} = \frac{27/(2\sqrt5)}{9/(2\sqrt5)} = 3$$

Solutions · 1