Question

which point could be on the line that is parallel to line kl and passes through point m?
\\( \bigcirc \\) (-10, 0)
\\( \bigcirc \\) (-6, 2)
\\( \bigcirc \\) (0, -6)
\\( \bigcirc \\) (8, -10)

Explanation:

Step1: Find slope of line KL

Points \( K(-6, 8) \) and \( L(6, 0) \). Slope \( m_{KL}=\frac{0 - 8}{6 - (-6)}=\frac{-8}{12}=-\frac{2}{3} \).

Step2: Equation of line through M

Point \( M(-4, -2) \), slope \( m = -\frac{2}{3} \) (parallel lines have equal slopes). Using point - slope form \( y - y_1 = m(x - x_1) \):
\( y - (-2)=-\frac{2}{3}(x - (-4)) \)
\( y + 2=-\frac{2}{3}(x + 4) \)
\( y=-\frac{2}{3}x-\frac{8}{3}-2 \)
\( y=-\frac{2}{3}x-\frac{14}{3} \)

Step3: Test each option

• For \( (-10,0) \): \( y=-\frac{2}{3}(-10)-\frac{14}{3}=\frac{20 - 14}{3}=2

eq0 \)

• For \( (-6,2) \): \( y=-\frac{2}{3}(-6)-\frac{14}{3}=\frac{12 - 14}{3}=-\frac{2}{3}

eq2 \)

• For \( (0,-6) \): \( y=-\frac{2}{3}(0)-\frac{14}{3}=-\frac{14}{3}

eq - 6 \)

• For \( (8,-10) \): \( y=-\frac{2}{3}(8)-\frac{14}{3}=\frac{-16 - 14}{3}=\frac{-30}{3}=-10 \) (matches)

Answer:

\( (8, -10) \)