Question

The lines ${(lx + my)^2} - 3{(mx - ly)^2} = 0$ and $lx + my + n = 0$ forms
A. an isosceles triangle
B. a right angled triangle
C. an equilateral triangle
D. None of the above

Answer 1

Hint: To find which kind of triangle will form first we will simplify the given equation to the simplest form so that we can easily compare it with the general equation. After comparing we will find the value of $a$, $b$ and $h$. Then put all these values in the formula for the angle between lines. With the help of angle, we will easily identify the kind of triangle.

Formula Used:
 $\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$

Complete step by step solution: Given, equation ${(lx + my)^2} - 3{(mx - ly)^2} = 0$ and $lx + my + n = 0$
${(lx + my)^2} - 3{(mx - ly)^2} = 0$
We know ${(a \pm b)^2} = {a^2} \pm 2ab + {b^2}$
${l^2}{x^2} + {m^2}{y^2} + 2lmxy - 3({m^2}{x^2} + {l^2}{y^2} - 2lmxy)$
${l^2}{x^2} + {m^2}{y^2} + 2lmxy - 3{m^2}{x^2} - 3{l^2}{y^2} + 6lmxy$
After simplifying, we get
$({l^2} - 3{m^2}){x^2} + ({m^2} - 3{l^2}){y^2} + 8lmxy = 0$
The general equation is $a{x^2} + 2hxy + b{y^2} = 0$
On comparing, we will get
$a = ({l^2} - 3{m^2})$, $h = 4lm$ and $b = {m^2} - 3{l^2}$
We know the formula for angle between the two lines is
$\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Putting $a = ({l^2} - 3{m^2})$, $h = 4lm$ and $b = {m^2} - 3{l^2}$
$\tan \alpha = \left| {\dfrac{{2\sqrt {{{\left( {4lm} \right)}^2} - ({l^2} - 3{m^2})({m^2} - 3{l^2})} }}{{{l^2} - 3{m^2} + {m^2} - 3{l^2}}}} \right|$
$\tan \alpha = \left| {\dfrac{{2\sqrt {16{l^2}{m^2} - \left( {{l^2}{m^2} - 3{l^4} - 3{m^4} + 9{m^2}{l^2}} \right)} }}{{ - 2{m^2} - 2{l^2}}}} \right|$
$\tan \alpha = \left| {\dfrac{{2\sqrt {16{l^2}{m^2} - \left( {10{l^2}{m^2} - 3({l^4} + {m^4})} \right)} }}{{ - 2{m^2} - 2{l^2}}}} \right|$
After simplification, we get
$\tan \alpha = \left| {\dfrac{{2\sqrt {16{l^2}{m^2} - 10{l^2}{m^2} + 3({l^4} + {m^4})} }}{{ - 2({m^2} + {l^2})}}} \right|$
$\tan \alpha = \left| {\dfrac{{2\sqrt {6{l^2}{m^2} + 3({l^4} + {m^4})} }}{{ - 2({m^2} + {l^2})}}} \right|$
After simplification, we get
$\tan \alpha = \dfrac{{\sqrt {6{l^2}{m^2} + 3{l^4} + 3{m^4}} }}{{({m^2} + {l^2})}}$
$\tan \alpha = \dfrac{{\sqrt 3 \sqrt {2{l^2}{m^2} + {l^4} + {m^4}} }}{{({m^2} + {l^2})}}$
$\tan \alpha = \dfrac{{\sqrt 3 \sqrt {{{\left( {{m^2} + {l^2}} \right)}^2}} }}{{({m^2} + {l^2})}}$
$\tan \alpha = \dfrac{{\sqrt 3 \left( {{m^2} + {l^2}} \right)}}{{\left( {{m^2} + {l^2}} \right)}}$
$\tan \alpha = \sqrt 3 $
Taking ${\tan ^{ - 1}}$ on both side
${\tan ^{ - 1}}\left( {\tan \alpha } \right) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right)$
We ${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \dfrac{\pi }{3}$
$\alpha = \dfrac{\pi }{3}$
Hence, the triangle is an equilateral triangle.

Therefore, option C is correct

Note: Students should simplify the given equations correctly as it is a complicated equation they can make mistakes while solving that. They should know the value of ${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \dfrac{\pi }{3}$ because without them they will not be able to solve the question.