Question

Two straight lines cut the axis of $x$ at distances $a$ and $-a$ and the axis of $y$ at distances $b$ and $b^{\prime }$ respectively; find the coordinates of their point of intersection.

Answer 1

Given straight cuts $x$ axis at $a$ and $y$ axis at $b$ , so it passes through $(a,0)$ and $(0,b)$

Equation of line is

$y-0=\frac{b-0}{0-a}(x-a)y-0=-\frac{b}{a}(x-a)ay=-bx+abay+bx=ab......\left(i\right)$

The other straight line passes through $(-a,0)$ and $(0,b^{\prime })$ , so its equation is

$y-0=\frac{b^{\prime }-0}{0-(-a)}(x-(-a\left)\right)y=\frac{b^{\prime }}{a}(x+a)$

substituting $y$ in $\left(i\right)$

$a\left(\frac{b^{\prime }x+a{b}^{\prime }}{a}\right)+bx=ab\frac{a{b}^{\prime }x+a^2{b}^{\prime }+abx}{a}=ab(a{b}^{\prime }+ab)x+a^2{b}^{\prime }=a^{2}ba(b+b^{\prime })x=a^2(b-b^{\prime })\Rightarrow x=\frac{a(b-b^{\prime })}{b+b^{\prime }}y=\frac{b^{\prime }}{a}(\frac{a(b-b^{\prime })}{b+b^{\prime }}+a)y=\frac{b^{\prime }}{a}\left(\frac{ab-a{b}^{\prime }+ab+a{b}^{\prime }}{b+b^{\prime }}\right)y=\frac{b^{\prime }}{a}\left(\frac{2ab}{b+b^{\prime }}\right)\Rightarrow y=\frac{2b{b}^{\prime }}{b+b^{\prime }}$

So the point of intersection is $(\frac{a(b-b^{\prime })}{b+b^{\prime }},\frac{2b{b}^{\prime }}{b+b^{\prime }})$