Question

Please help me out. I don’t understand this problem.. An equation of a line through (-7,-10) which is perpendicular to the line y=-3x+2 has a slope ____ and y-intercept at ___...

Answer 1

Answer:

The Equation:

$y = \frac{1}{3}x - \frac{23}{3}$

The y-intercept:

$-\frac{23}{3}$

Step-by-step explanation:

Things you need to know:

Perpendicular mean opposite slope.

Need to know point slope form of a linear equation $y - y_1 = x(x - x_1)$

Ok, with that out the way, lets get started!

From the equation y = -3x + 2 we can see the slope, which is -3. We are asked to find an equation that is perpendicular to that slope so that means we need an opposite slope. To turn this slope to an opposite slope flip it and change the sign

-3 = $\frac{-3}{1} = \frac{1}{3}$

Now that we have our opposite slope, which is $\frac{1}{3}$, we can find the equation the problem is asking us to solve.

We do this by using the point (-7, -10) and our new slope, which is $\frac{1}{3}$ and by using  $y - y_1 = x(x - x_1)$

Insert our point and slope into $y - y_1 = x(x - x_1)$ and solve for y

$y - y_1 = x(x - x_1)$

$y - (-10) = \frac{1}{3}(x - (-7))$

$y + 10) = \frac{1}{3}(x + 7))$

Work on the right side of the = sign

Distribute 1/3 to x and 7

$y + 10 = \frac{1}{3}(x + 7))$

$y + 10 = \frac{1}{3}x + \frac{7}{3}$

Bring the 10 over to the right side by adding a -10

$y + 10 -10= \frac{1}{3}x + \frac{7}{3} - 10$

$y = \frac{1}{3}x + \frac{7}{3} - 10$

Do the math for $\frac{7}{3} - 10 = -\frac{23}{3}$

$y = \frac{1}{3}x - \frac{23}{3}$

We are done.

The Equation:

$y = \frac{1}{3}x - \frac{23}{3}$

The y-intercept:

$-\frac{23}{3}$