Question

question 3(multiple choice worth 1 points) (04.02 mc) square efgh is drawn on a coordinate plane. diagonal fh is on the line y - 3 = -\frac{1}{3}(x + 9). what is the slope of the diagonal ge? -\frac{1}{3} \frac{1}{3} -3 3

Explanation:

Step1: Identify slope of FH

The equation of line FH is in point - slope form $y - y_1=m(x - x_1)$, where $m =-\frac{1}{3}$ is the slope of line FH.

Step2: Recall property of square diagonals

In a square, the diagonals are perpendicular to each other. If the slope of one line is $m_1$ and the slope of a perpendicular line is $m_2$, then $m_1\times m_2=- 1$.

Step3: Calculate slope of GE

Let the slope of FH be $m_1 =-\frac{1}{3}$ and the slope of GE be $m_2$. We have $-\frac{1}{3}\times m_2=-1$. Solving for $m_2$ gives $m_2 = 3$.

Answer:

D. 3