Question

Find the equation of the line that satisfies the given conditions.

Through $\left(-2,-11\right)$; perpendicular to the line passing through $\left(1,1\right)$and $\left(5,-1\right)$

Short answer

Equation of the line is $2x-y-7=0$.

Step-by-step solution

Step 1. Definition of slope of the line passing through two points and point-slope form.

• Slope of the line is$m=\frac{y_2-y_1}{x_2-x_1}$, where $\left(x_1,y_1\right),\left(x_2,y_2\right)$are coordinates of points lying on the line.
• Point-slope form is the equation of a straight line in the form $y-y_1=m\left(x-x_1\right)$, where m is the slope of the line and$\left(x_1,y_1\right)$ are the coordinates of the given point on the line.

Step 2. Calculate the slope.

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

$\left(x_1,y_1\right):\left(1,1\right)$

$\left(x_2,y_2\right):\left(5,-1\right)$

$\begin{array}{l}m=\frac{-1-1}{5-1}\\ m=\frac{-2}{4}\\ m=-\frac{1}{2}\end{array}$

$\begin{array}{l}{m}_{\text{perpendicular}}=-\left(\frac{-2}{1}\right)\\ =2\end{array}$

Step 3. Write the equation in general form.

Use the point-slope form to write the equation of the line parallel to the given line:

Use $x_1=-2,y_1=-11$, and the slope is$m=2$

$\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-\left(-11\right)=2\left(x-\left(-2\right)\right)\\ y+11=2\left(x+2\right)\\ y+11=2x+4\\ 2x+4-y-11=0\\ 2x-y-7=0\end{array}$