# What is the equation of the line that passes through (-1,1)  and is perpendicular to the line that passes through the following points: (-2,4),(-12,21) ?

Mar 11, 2018

$y = \frac{10}{17} x + \frac{27}{17}$

#### Explanation:

Consider the two points $\left(- 2 , 4\right) \mathmr{and} \left(- 12 , 21\right)$

The straight line between them has the gradient:

$m = \left(\text{change in "y)/("change in } x\right)$ reading left to right on the x-axis

$m = \frac{21 - 4}{- 12 - \left(- 2\right)} = - \frac{17}{10}$

The gradient of the perpendicular line is $- \frac{1}{m} = + \frac{10}{17}$

So we have $y = \frac{10}{17} x + c$ passing through $\left(x , y\right) \to \left(- 1 , 1\right)$

So by substitution we get:

$y = \frac{10}{17} x + c \textcolor{w h i t e}{\text{dddd")->color(white)("dddd}} 1 = \frac{10}{17} \left(- 1\right) + c$

$c = 1 + \frac{10}{17} = \frac{27}{17}$

$y = \frac{10}{17} x + \frac{27}{17}$

Mar 11, 2018

Equation of line is $10 x - 17 y = - 27$.

#### Explanation:

The slope of the line passing through $\left(- 2 , 4\right) \mathmr{and} \left(- 12 , 21\right)$

is ${m}_{1} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{21 - 4}{- 12 + 2} = - \frac{17}{10}$

The product of slopes of the pependicular lines is ${m}_{1} \cdot {m}_{2} = - 1$

$\therefore {m}_{2} = - \frac{1}{- \frac{17}{10}} = \frac{10}{17}$. Hence the slope of the pependicular

line is ${m}_{2} = \frac{10}{17}$ . Equation of line passing through $\left(- 1 , 1\right)$

is $y - {y}_{1} = {m}_{2} \left(x - {x}_{1}\right) \mathmr{and} y - 1 = \frac{10}{17} \left(x + 1\right)$ or

$17 y - 17 = 10 \left(x + 1\right) \mathmr{and} 10 x - 17 y = - 27$.

Equation of line is $10 x - 17 y = - 27$. [Ans]