Question

72. If the pair of straight lines given by ax* + 2hxy + by2 = 0 are perpendicular then
(a) ab=0
(b) a+b=0
(c) a-b=0
(d) a=0​

Answer 1

First let's find the individual lines.

So we're given the pair of straight lines,

$ax^2+2hxy+by^2=0$

We are going to factorise this equation, by splitting the middle term. Or you can use quadratic method to solve it. So,

$ax^2+2hxy+by^2=0\\\\\\ax^2+\left (\dfrac {2h+\sqrt {4h^2-4ab}}{2}\right)xy+\left (\dfrac {2h-\sqrt {4h^2-4ab}}{2}\right)xy+by^2=0\\\\\\ax^2+\left(h+\sqrt {h^2-ab}\right)xy+\left(h-\sqrt {h^2-ab}\right)xy+by^2=0\\\\\\x^2+\left (\dfrac {h+\sqrt {h^2-ab}}{a}\right)xy+\left (\dfrac {h-\sqrt {h^2-ab}}{a}\right)xy+\left (\dfrac {b}{a}\right)\ y^2=0\\\\\\\left (x+\left (\dfrac {h+\sqrt {h^2-ab}}{a}\right)y\right)\left (x+\left (\dfrac {h-\sqrt {h^2-ab}}{a}\right)y\right)=0$

$\left (\dfrac {ax+(h+\sqrt {h^2-ab})y}{a}\right)\left (\dfrac {ax+(h-\sqrt {h^2-ab})y}{a}\right)=0$

This implies,

$(ax+(h+\sqrt {h^2-ab})y)(ax+(h-\sqrt {h^2-ab})y)=0$

So the individual lines are,

$ax+(h+\sqrt {h^2-ab})y=0\\\\ax+(h-\sqrt {h^2-ab})y=0$

The lines meet each other at the origin, that's why they don't have constant terms.

We know that slope of a line $Ax+By+C=0$ is $-\dfrac {A}{B}.$ So,

Slope of the line $ax+(h+\sqrt {h^2-ab})y=0$ is,

$-\dfrac {a}{h+\sqrt {h^2-ab}}.$

Similarly,

Slope of the line $ax+(h-\sqrt {h^2-ab})y=0$ is,

$-\dfrac {a}{h-\sqrt {h^2-ab}}.$

For these lines being perpendicular, one should be the negative reciprocal of the other. Thus,

$-\dfrac {a}{h+\sqrt {h^2-ab}}=\dfrac {h-\sqrt {h^2-ab}}{a}\\\\\\(h+\sqrt {h^2-ab})(h-\sqrt {h^2-ab})=-a^2\\\\\\h^2-(h^2-ab)=-a^2\\\\\\ab=-a^2\\\\\\a^2+ab=0\\\\\\a(a+b)=0$

Since $a\neq0,$

$\Large\boxed {\mathbf {a+b=0}}$

Hence $\mathbf {(b)}$ is the answer.