Question

Equations $3x^2-8xy-3y^2=0$ and $x-2y=3$ represents the sides of a triangle which is

• A. right angled but not isosceles
• B. only isosceles
• C. equilateral
• D. right angled and isosceles

Answer 1

The correct option is A right angled but not isosceles

Given eq of pair of line

$3x^2-8xy-3y^2=0$

and line $x-2y=3$

Here condition to be perpendicular of 2 lines of pair $a{x}^2+2hxy+b{y}^2=0$ is

$a+b=0$

$3-3=0\Rightarrow 0$

hence line is perpendicular SO the lines making right angled triangle

Equation of angle of bisector

$\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

$\frac{x^2-y^2}{3+3}=\frac{xy}{-4}$

$\frac{x^2-y^2}{6}=\frac{xy}{-4}$

the lines from above eq are

$y=2x$ and $y=-\frac{1}{2}x$

so both above eq is not perpendicular to $x-2y=3$

$\therefore$ the triangle is not isosceles