Question

What is the slope of the line perpendicular to the line passing through the points (-2, 3) and (2, 0?

Answer 1

$\huge \boxed{\mathcal{ANSWER}}$

$\mathtt{Let \: the\: points\:be \:A (-2,3)\: and\: B(2,0)}$

$\mathtt{When \: we \: have \: to \: find \: slope \: of}$ $\mathtt{ line \: passing \: through \: two \: fixed \: points}$
$\mathtt{ Such\:that\: coordinates \: of \:those \: two \:points \:are}$

$\bold{P(x_1,y_1) \: and Q(x_2,y_2)}$

$\boxed{Slope \: m = \frac{y_1 - y_2}{x_1 - x_2}}$

or,

$\boxed{Slope \: m = \frac{Difference \: of \: ordinates \: of \: the \: given \: point}{Diffrence \: in \: their \: abcissea }}$

$\mathtt{On \: applying \: the \: values}$

$\boxed{Slope \: m = \frac{3 - 0}{(-2) - (2)}}$

$\boxed{Slope \: m = \frac{3 }{-4}}$

NOW,

$\mathtt{Let \: the\: slope\: of\: second\: line \:be \:m_2}$

$\mathtt{Since,\: the \: given\:second\: line \:is \: perpendicular \: to }$$\mathtt{first \: line \: so \: their \: product \: will \: be \: one }$

$\boxed{m \times m_2 =1}$

$\mathtt{So,\: on \: placing \: value \: of \: first \:line }$

$\bold {\frac{3}{-4} \times m_2 =1}$

$\mathtt{On \: cross \: multiplication \: we \: get}$

$\bold { 3\times m_2 =1 \times -4}$

$\bold { m_2 =\frac{-4}{3}}$

$\mathtt{So\: the \: slope \:of \: line \: perpendicular \: to }$$\mathtt{first \: line \: is \frac{-4}{3}}$