Q25. - 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.

$\left(-1,-4\right);-x-2y=0$

Step-by-Step Solution

Answer

The equation in the slope-intercept form for the line that passes through the point $\left(-1,-4\right)$ and is perpendicular to the given equation is $\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{2}$.

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 $-x-2y=0$. Convert the given equation in the standard slope and intercept form of the equation $y=mx+b$.

$$\begin{gathered} -x-2y=0 \\ 2y=-x \\ y=\frac{-1}{2}x \end{gathered}$$

Compared to the standard form of the equation $y=mx+b$, the slope of the equation is:

 $m=\frac{-1}{2}$

Calculate the slope of the line which is perpendicular to the given line $-x-2y=0$.

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

 

 

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 $\left(-1,-4\right)$ and is perpendicular to the equation $-x-2y=0$, 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 $x_1=-1,y_1=-4$.

 $$\begin{gathered} y-y_1=m\left(x-x_1\right) \\ y-\left(-4\right)=2\left(x-\left(-1\right)\right) \\ y+4=2x+2 \\ y=2x-2 \end{gathered}$$

Therefore, the equation in the slope-intercept form for the line that passes through the point $\left(-1,-4\right)$ and is perpendicular to the given equation is $y=2x-2$.