Question

Without finding the vertices or angles of the triangle, show that the three straight lines $au+bv=0$; $au-bv=2ab$ and $u+b=0$ from an isosceles triangle where $u=x+y-b$ and $v=x-y-a$ and $a,b\ne 0$

Answer 1

REF.Image

Here the slope of $au+bv=0$ is $(-\frac{a}{b})$

& the slope of $av-bv=2ab$ is $\left(\frac{a}{b}\right)$

Hence both of the linear are moving

same angle in apposite direction

with v-axis & $u+b=0$ is parallel to v-axis

hence, these two lines are moving same

angle with $u+b=0$ from their point of intersection

Hence $△ABC$ is an isosceles triangle