Question

line dg is represented by the equation y=-x + 11. determine the equation, in slope - intercept form, of the line qf that is perpendicular to line dg and passes through the point q(8,19). y =
slope of line dg|slope of line qf|point - slope form of line qf
m1|m2|y - y1 = m(x - x1)

Explanation:

Step1: Find slope of line DG

The equation of line DG is $y=-x + 11$, so the slope $m_1=-1$.

Step2: Find slope of line QF

For two perpendicular lines, the product of their slopes is - 1. Let the slope of line QF be $m_2$. Since $m_1\times m_2=-1$ and $m_1 = - 1$, then $(-1)\times m_2=-1$, so $m_2 = 1$.

Step3: Use point - slope form to find equation of line QF

The line QF passes through the point $(x_1,y_1)=(8,19)$ and has slope $m = 1$. The point - slope form is $y - y_1=m(x - x_1)$. Substituting the values, we get $y-19=1\times(x - 8)$.

Step4: Convert to slope - intercept form

Expand the point - slope form: $y-19=x - 8$. Then, add 19 to both sides to get $y=x+11$.

Answer:

$y=x + 11$