Question

Find the equation of the line that satisfies the given conditions.(Lesson 1.10)

Through$\left(2,1\right)$and$\left(1,6\right)$

Short answer

Equation of the line is $5x+y-11=0$.

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.

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

$\left(x_1,y_1\right):\left(2,1\right)$

$\left(x_2,y_2\right):\left(1,6\right)$

$\begin{array}{l}m=\frac{6-1}{1-2}\\ m=\frac{5}{-1}\\ m=-5\end{array}$

Step 3. Write the equation in general form.

Use the point-slope form to write the equation of the line with $x_1=2$, $y_1=1$, and the slope is$m=-5$

$\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-1=-5\left(x-2\right)\\ y-1=-5x+10\\ y-1+5x-10=0\\ 5x+y-11=0\end{array}$