Question

line se is represented by the equation y=-x - 13. determine the equation, in slope - intercept form, of the line gp that is perpendicular to line se and passes through the point g(-4,-10).
slope of line se: m1
slope of line gp: m2
point - slope form of line gp: y - y1=m(x - x1)

Explanation:

Step1: Find slope of line SE

The equation of line SE is $y=-x - 13$, which is in the slope - intercept form $y = mx + b$ where $m$ is the slope. So, the slope of line SE, $m_1=-1$.

Step2: Find slope of line GP

If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of line GP 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: Write point - slope form of line GP

The point - slope form of a line is $y - y_1=m(x - x_1)$. The line GP passes through the point $G(-4,-10)$ and has a slope $m_2 = 1$. Substituting $x_1=-4$, $y_1=-10$ and $m = 1$ into the point - slope form, we get $y-(-10)=1\times(x - (-4))$, which simplifies to $y + 10=x + 4$.

Answer:

Slope of line SE ($m_1$): $-1$
Slope of line GP ($m_2$): $1$
Point - slope form of line GP: $y + 10=x + 4$