Q23. - Algebra 1

Question

Write an equation in slope-intercept form for the line that passes through the given point and is perpendicular to the graph of the equation.

$\left(0,-3\right);y=-2x+4$

Step-by-Step Solution

Answer

The equation in the slope-intercept form for the line that passes through the point $\left(0,-3\right)$ and is perpendicular to the given equation is $\mathit{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{x}\mathbf{-}\mathbf{3}$.

Step 1. State the concept of slope-intercept of an equation and slope of parallel and perpendicular lines.

The slope-intercept form of a linear equation is $y=mx+b$, where m is the slope and $b$ is the $y$-intercept.

Parallel lines: Slopes $m_1$ and $m_2$ of the parallel lines are same i.e., $m_1=m_2$.

Perpendicular lines: The slopes $m_1$ and $m_2$ of the nonvertical perpendicular lines are opposite reciprocals i.e., $m_1{m}_2=-1$. 

Step 2. Calculate the slope.

Calculate the slope of the first equation $y=-2x+4$. The equation is in the form of slope and intercept form. Compared from the standard form of the equation $y=mx+b$, the slope of the equation is:

$m=-2$

Calculate the slope of the line which is perpendicular to the given line $y=-2x+4$.

$$\begin{gathered} m_1{m}_2=-1 \\ \left(-2\right)m_2=-1 \\ m_2=\frac{1}{2} \end{gathered}$$

Step 3. Write a new equation in slope-intercept form.

To find the equation in slope-intercept form for the line that passes through the point $\left(0,-3\right)$ and is perpendicular to the equation $y=-2x+4$, substitute the value of slope and point in the standard form of slope- point form of the equation i.e., $y-y_1=m\left(x-x_1\right)$ where in $x_1=0,y_1=-3$. 

 $$\begin{gathered} y-y_1=m\left(x-x_1\right) \\ y-\left(-3\right)=\frac{1}{2}\left(x-0\right) \\ y+3=\frac{1}{2}x \\ y=\frac{1}{2}x-3 \end{gathered}$$

Therefore, the equation in the slope-intercept form for the line that passes through the point $\left(0,-3\right)$ and is perpendicular to the given equation is $y=\frac{1}{2}x-3$.