Q 13. - Precalculus Enhanced with Graphing Utilities

Question

For the line $2x+3y=6$, find a line parallel to it containing the point $\left(1,-1\right)$.Also find a line perpendicular to it containing the point $\left(0,3\right)$.

Step-by-Step Solution

Answer

Line parallel to $2x+3y=6$ this line is $2x+3y=-1$and the line perpendicular to this line is  $2x+3y=-9$.
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Step 1. Given information.

Consider the given line $2x+3y=6$ and points $\left(1,-1\right)$ and $\left(0,3\right)$.

Step 2. Compare the equation with the slope intercept form of the equation of line.

Equation of slope intercept form is $y=mx+b$ where m is slope and $b$is $y$ intercept.

$$\begin{gathered} 2x+3y=6 \\ 3y=-2x+6 \\ y=\frac{-2}{3}x+2 \end{gathered}$$

For parallel lines $m_1=m_2$ where $m_1=\frac{-2}{3}$

Further simplify to find $m_2$.

Step 3. To use point slope form of equation of line.

Point slope form of equation of line is $y-y_1=m\left(x-x_1\right)$.

Substitute the value of $x_1=1 ,y_1 -1$and $m=\frac{-2}{3}$.

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

Solving this equation in the form of $y=mx+b$we get the value of $m_2=\frac{-2}{3}$.

Therefore $m_1=m_2$ so the lines are parallel.

Step 4. Formula and slope property used to show that both the lines are perpendicular.

Formula used to find the equation of line is $y-y_1=m\left(x-x_1\right)$ and for two lines to be perpendicular $m_1{m}_2=-1$.

Substitute the value of $x_1=0 ,y_1=3$ Therefore the equation of line is $2x+3y=-9$ .So both the  lines are perpendicular .