Answer
a) MQ has length $m_1$.
b) QN has length $-m_2$.
c) Triangles MPQ and PNQ are similar.
d) $m_1/1$=$1/(-m_2)$ and $m_1m_2$=-1
Work Step by Step
a) We can assume without loss of generality that the point P is at the origin $(0,0)$. The coordinates of Q is $(1,0)$. Since the line $L_1$ intersects y-axis at 0, $b_1$=0. The equation of the line $L_1$ is, therefore, $y=m_1x$. The coordinates of M is ($m_1x,x$) at x=1, i.e. ($m_1,1$). By the distance formula, |MQ|=$\sqrt{m_1^2+0^2}$=$|m_1|=m_1$ since $m_1>0$.
b) Similar to part a), equation to line $L_2$ is $y= m_2x$ and the coordinate of N is $(m_2,1) $. By, distance formula, |NQ|=$\sqrt {m_2^2+0^2}=|m_2|=-m_2$ since $m_2<0$.
c) Consider $\triangle MPQ$ & $\triangle PNQ$.
i) $\angle MQP=\angle NQP=90\circ$
Let $\angle QPN=\alpha$
ii) $\angle MPQ=\angle PNQ = 90-\alpha $ (By the angle sum property of $\triangle PQN$)
iii) $\angle PMQ=\angle QPN=\alpha$ (By the angle sum property of $\triangle
PMQ$)
BY AAA similarity, $\triangle MPQ \sim \triangle PNQ$
d) Consider $\triangle PMQ$ and $\triangle PNQ$, By Pythagorean theorem,
$PM^2=1+m_1^2$
$PN^2=1+m_2^2$
Now, applying Pythagorean theorem in the right $\triangle PMN$,
$MN^2=PM^2+PN^2$
$(m_1-m_2)^2=1+m_1^2+1+m_2^2$
$m_1^2-2m_1m_2+m_2^2=m_1^2+m_2^2+2$
$m_1m_2=-1$
$\implies m_1/1=1/(-m_2)$
Hence Proved.
