Question

what is the equation of line passing through (-1, 2) and perpendicular to x-y+2=0? ​

Answer 1

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Q1) What is the equation of line passing through (-1,2) and perpendicular to x-y+2=0

$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

$\: \large \orange{ \bold{ \underline{ step \: by \: step \: explaination : }}}$

$\pink{\frak{Given}\begin{cases} \sf equation \: of \: line \: passing \:through \: (-1,2). \\ \: perpendicular \: line \: x-y+2=0 \\ \: to \: find \to \: the\: equation \: of \: line =? \end{cases}}$

$\: \implies \displaystyle \sf \: x - y + 2 = 0$

$\: \implies \displaystyle \sf - y + x + 2 = 0$

$\: \implies \displaystyle \sf \: y - x - 2 = 0$

$\: \implies \displaystyle \sf \: y = x + 2$

$\: \quad \diamondsuit\underline{ \bf{ \: slope \: of \: line \: is \: 1}}$

$\: \: \implies \displaystyle \sf \: (y2 - y1) = m(x2 - x1)$

$\: \: \implies \displaystyle \sf{(y2 - 2) = 1(x2 + 1)}$

$\: \: \implies \displaystyle \sf \: (y - 2) = 1(x + 1)$

$\: \: \implies \displaystyle \sf \: y - 2 = x + 1$

$\: \implies \displaystyle \sf \: y - 2 - x + 1 = 0$

$\: \: \implies \displaystyle \sf \: - x + y - 1 = 0$

$\: \implies \displaystyle \sf \: x - y + 1 = 0$

$\: \boxed{ \boxed{ \underline{ \bf{the \:equation \: of \: line \: is \: x - y + 1 = 0}}}}$