Chapter 7 - Cumulative Review Exercises - Page 880: 32

The linear equation in slope-intercept form is $y=3x+3$.

Work Step by Step

We know that if two lines are perpendicular, then the product of their slopes must be $-1$. Considering that ${{m}_{1}}$ and ${{m}_{2}}$ are slopes, the condition of perpendicularity is: ${{m}_{1}}\times {{m}_{2}}=-1$ Therefore, the slope of the line $x+3y-6=0$ can be computed as: \begin{align} & x+3y-6=0 \\ & 3y=-x+6 \\ & y=-\frac{x}{3}+2 \end{align} So, the slope of the line is ${{m}_{1}}=-\frac{1}{3}$. And, \begin{align} & -\frac{1}{3}\times {{m}_{2}}=-1 \\ & {{m}_{2}}=3 \end{align} Therefore, the equation of the line is written as given below: \begin{align} & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & y-0=3\left( x+1 \right) \\ & y=3x+3 \end{align} Thus, the linear equation in slope-intercept form is $y=3x+3$.

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