# How do you find the slope and y intercept of the line that is perpendicular to y=-x-1 and passes through the point (5, 7)?

Nov 26, 2016

The equation of the line is $y = x + 2$
The slope is $= 1$
The intercepts are $\left(0 , 2\right)$ and $\left(- 2 , 0\right)$

#### Explanation:

The equation of a line is $y = m x + c$

where $m$ is the slope

The slope of the line $y = - x - 1$ is ${m}_{1} = - 1$

The slope of the perpedicular line is ${m}_{2}$

${m}_{1} \cdot {m}_{2} = - 1$

So ${m}_{2} = 1$

The equation of a line, that passes through $\left({x}_{0} , {y}_{0}\right)$ with a slope of ${m}_{2}$ is

$y - {y}_{0} = {m}_{2} \left(x - {x}_{0}\right)$

Here, $\left({x}_{0} , {y}_{0}\right) = \left(5 , 7\right)$

So, the equation is

$y - 7 = 1 \left(x - 5\right)$

$y = x + 2$

graph{(y+x+1)(y-x-2)=0 [-7.9, 7.9, -3.95, 3.95]}