Question

x/a = y/b ; ax + by = a^2 + b^2 solve by elimination or cross multiplication method​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of lines are

$\rm :\longmapsto\:\dfrac{x}{a} = \dfrac{y}{b}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{x}{a} - \dfrac{y}{b} = 0$

$\rm :\longmapsto\:\dfrac{bx - ay}{a} = 0$

$\rm :\longmapsto\:bx - ay = 0 - - - (1)$

and

$\rm :\longmapsto\:ax + by = {a}^{2} + {b}^{2} - - - (2)$

Now,

Using Cross Multiplication Method

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf - a & \sf 0 & \sf b & \sf - a\\ \\ \sf b & \sf {a}^{2} + {b}^{2} & \sf a & \sf b\\ \end{array}} \\ \end{gathered}$

So,

$\rm :\longmapsto\:\dfrac{x}{ - a( {a}^{2} + {b}^{2}) - 0 } = \dfrac{y}{0 - b( {a}^{2} + {b}^{2})} = \dfrac{ - 1}{ {b}^{2}+{a}^{2} }$

$\rm :\longmapsto\:\dfrac{x}{ - a( {a}^{2} + {b}^{2})} = \dfrac{y}{- b( {a}^{2} + {b}^{2})} = \dfrac{1}{ - ({b}^{2}+{a}^{2})}$

$\red{\rm :\longmapsto\:On \: multiply \: by \: - ({a}^{2} + {b}^{2}), \: we \: get}$

$\rm :\longmapsto\:\dfrac{x}{a} = \dfrac{y}{b} = 1$

$\bf\implies \:x = a \: \: \: or \: \: \: y = b$

Verification :-

Consider Equation (1),

$\rm :\longmapsto\:bx - ay = 0$

On substituting the values of x and y, we get

$\rm :\longmapsto\:ba - ab = 0$

$\rm :\longmapsto\:0 = 0$

Hence, Verified

Consider Equation (2),

$\rm :\longmapsto\:ax + by = {a}^{2} + {b}^{2}$

On substituting the values of x and y, we have

$\rm :\longmapsto\:a(a) + b(b) = {a}^{2} + {b}^{2}$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} = {a}^{2} + {b}^{2}$

Hence, Verified