Q22. - 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.

 $\left(3,-4\right);y=-\frac{1}{3}x-5$

Step-by-Step Solution

Answer

The equation in the slope-intercept form for the line that passes through the point $\left(3,-4\right)$ and is perpendicular to the given equation is $\mathit{y}\mathbf{=}\mathbf{3}\mathit{x}\mathbf{-}\mathbf{13}$.

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 $y=-\frac{1}{3}x-5$. The equation is in the form of slope and intercept form. Compared to the standard form of the equation $y=mx+b$, the slope of the equation is:

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

Calculate the slope of the line which is perpendicular to the given line $y=-\frac{1}{3}x-5$.

  $$\begin{gathered} m_1{m}_2=-1 \\ \left(\frac{-1}{3}\right)m_2=-1 \\ m_2=3 \end{gathered}$$

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 $\left(3,-4\right)$ and is perpendicular to the equation $y=-\frac{1}{3}x-5$, 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 $x_1=3,y_1=-4$.

$$\begin{gathered} y-y_1=m\left(x-x_1\right) \\ y-\left(-4\right)=3\left(x-3\right) \\ y+4=3x-9 \\ y=3x-13 \end{gathered}$$

Therefore, the equation in the slope-intercept form for the line that passes through the point $\left(3,-4\right)$ and is perpendicular to the given equation is $y=3x-13$.