What is the equation of the line that passes through (-2,7) and is perpendicular to the line that passes through the following points: (1,3),(-5,6) ?

May 25, 2018

color (purple)(2x - y + 11 = 0

Explanation:

$A \left(1 , 3\right) , B \left(- 5 , 6\right) , C \left(- 2 , 7\right)$

Slope of $\vec{A B} = m = \frac{{y}_{b} - {y}_{a}}{{x}_{b} - {x}_{a}}$

$m = \frac{6 - 3}{- 5 - 1} = - \left(\frac{1}{2}\right)$

Slope of perpendicular line $= - \left(\frac{1}{m}\right) = 2$

Equation of line passing through C and perpendicular to $\vec{A B}$ is

$y - {y}_{c} = - \left(\frac{1}{m}\right) \cdot \left(x - {x}_{c}\right)$

$y - 7 = 2 \cdot \left(x + 2\right)$

$y - 7 = 2 x + 4$

color (purple)(2x - y + 11 = 0

May 25, 2018

$y = 2 x + 11$

Explanation:

$\text{first find the slope m of the line passing through the}$
$\text{the 2 given points}$

$\text{to calculate the slope m use the "color(blue)"gradient formula}$

•color(white)(x)m=(y_2-y_1)/(x_2-x_1)

$\text{let "(x_1,y_1)=(1,3)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 5 , 6\right)$

$m = \frac{6 - 3}{- 5 - 1} = \frac{3}{- 6} = - \frac{1}{2}$

$\text{given a line with slope m then the slope of a line}$
$\text{perpendicular to it is}$

•color(white)(x)m_(color(red)"perpendicular")=-1/m

"hence "m_("perpendicular")=-1/(-1/2)=2

$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.

•color(white)(x)y=mx+b

$\text{where m is the slope and b the y-intercept}$

$\text{here } m = 2$

$y = 2 x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$

$\text{to find b substitute "(-2,7)" into the partial equation}$

$7 = - 4 + b \Rightarrow b = 7 + 4 = 11$

$y = 2 x + 11 \leftarrow \textcolor{red}{\text{is the required equation}}$