Question

Write an equation of the line with the given conditions.

$\left(-3,-1\right)$; slope$\frac{1}{2}$.

Short answer

The equation of the straight line for the given condition is $\mathit{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{x}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}$.

Step-by-step solution

Step 1. State the concept of a straight line.

The standard form of slope and intercept form of the straight line is $y=mx+c$.

Here, $m$is the slope of the straight line and $c$is the $y$-intercept of the equation.

Step 2. State the equation of a line slope-point form.

The standard form of slope and point form of the straight line is$y-y_1=m\left(x-x_1\right)$

Here m is the slope of the straight line and is the point from which the straight line passes.

$\begin{array}{l}\left(x_1,y_1\right)=\left(-3,-1\right)\\ m=\frac{1}{2}\end{array}$

Step 3. Substitute the values to get the required equation.

Substitute the above values in the equation$y-y_1=m\left(x-x_1\right)$

$$\begin{gathered} y-y_1=m\left(x-x_1\right) \\ y-\left(-1\right)=\frac{1}{2}\left(x-\left(-3\right)\right) \\ y+1=\frac{1}{2}x+\frac{3}{2} \\ y+1-1=\frac{1}{2}x+\frac{3}{2}-1 \\ y=\frac{1}{2}x+\frac{1}{2} \end{gathered}$$

The equation of the straight line is $y=\frac{1}{2}x+\frac{1}{2}$.