Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=-\dfrac{1}{2}x-\dfrac{3}{2} \\\\\text{b) Standard Form: } x+2y=-3$
$\bf{\text{Solution Outline:}}$ Find the slope of the given equation, $y=2x .$ Then use the negative reciprocal of this slope and the given point, $(-7,2) ,$ to find the equation of the needed line. Give the equation in the Slope-Intercept Form and in the Standard Form. $\bf{\text{Solution Details:}}$ In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the given equation is equivalent to \begin{array}{l}\require{cancel} y=2x .\end{array} Hence, the slope of the given line is $m=2 .$ Since perpendicular lines have negative reciprocal slopes, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} m=-\dfrac{1}{2} \\\text{Through: } (-7,2) .\end{array} Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions, \begin{array}{l}\require{cancel} y_1=2 ,\\x_1=-7 ,\\m=-\dfrac{1}{2} ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-2=-\dfrac{1}{2}(x-(-7)) \\\\ y-2=-\dfrac{1}{2}(x+7) .\end{array} Using the properties of equality, in the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y-2=-\dfrac{1}{2}(x)-\dfrac{1}{2}(7) \\\\ y-2=-\dfrac{1}{2}x-\dfrac{7}{2} \\\\ y=-\dfrac{1}{2}x-\dfrac{7}{2}+2 \\\\ y=-\dfrac{1}{2}x-\dfrac{7}{2}+\dfrac{4}{2} \\\\ y=-\dfrac{1}{2}x-\dfrac{3}{2} .\end{array} Using the properties of equality, in the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y=-\dfrac{1}{2}x-\dfrac{3}{2} \\\\ 2(y)=2\left( -\dfrac{1}{2}x-\dfrac{3}{2} \right) \\\\ 2y=-x-3 \\\\ x+2y=-3 .\end{array} The equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=-\dfrac{1}{2}x-\dfrac{3}{2} \\\\\text{b) Standard Form: } x+2y=-3 .\end{array}