Question

Answer correctly above ques.​

Answer 1

$\huge \bold \color{red} \mathfrak{solution} :-$

$\bold{ \underline{ \underline{given \: lines : }}}$

$ax + 3y = 7 \\ and \\ (a - 1)x + 2y = b$

note:-

if general equation is in the form of ax+by+c = 0

then..slope

$\bold{slope = \frac{ - a}{b} }$

slope of line ax+3y=7 or ax+3y-7=0 :-

$\bold{ = \frac{ - a}{3} }$

slope of line (a-1)x+2y=b or (a-1)x+2y-b = 0 :-

$\bold{ = \frac{ - a + 1}{2} }$

we know that..if two lines are perpendicular..then.. product of their slopes is -1

$\bold{ \implies \frac{ - a}{3} \times \frac{( - a + 1)}{2} = - 1}$

$\bold{ \implies \frac{ {a}^{2} - a }{6} = - 1}$

$\bold{\implies {a}^{2} - a = - 6 }\\ \bold{\implies {a}^{2} - a + 6}$

now solve this quadratic equation by quadratic formula..

we get :-

$\bold{a = \frac{ - 1 - \sqrt{23} }{2} \: and \: \frac{ - 1 + \sqrt{23} }{2} }$

sum of squares of value of a :-

$\bold{ { (\frac{ - 1 - \sqrt{23} }{2} )}^{2} + {( \frac{1 + \sqrt{23} }{2} )}^{2} }$

$\bold{\implies( \frac{1 + 23 - 2 \sqrt{23} }{4} ) + ( \frac{1 + 23 + 2 \sqrt{23} }{4} )}$

$\bold{ \implies( \frac{12 - \sqrt{23} }{2} ) + ( \frac{12 + \sqrt{23} }{2})}$

$\bold{\implies \frac{12 - \sqrt{23 } + 12 + \sqrt{23} }{2} }$

$\bold{ \implies \frac{12 + 12}{2} } = 12$

so sum of squares of value of a is 12