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Question 11:

Triangle $KLM$ with vertices $K(0,4)$, $L(-3,-1)$, and $M(2,-2)$ is dilated with center $(-2,3)$ and scale factor 4. Find the coordinates of the new triangle $\Delta K'L'M'$.

The formula for dilation: $(x', y') = (a + k(x - a), b + k(y - b))$ Where:

• $(a, b)$ is the center of dilation $(-2, 3)$,
• $k = 4$ (scale factor),
• $(x, y)$ are the original coordinates.

Solution for each point:

1. For $K(0,4)$: $x' = -2 + 4(0 - (-2)) = -2 + 4(2) = 6$ $y' = 3 + 4(4 - 3) = 3 + 4(1) = 7$ New $K'(6,7)$.

2. For $L(-3,-1)$: $x' = -2 + 4(-3 - (-2)) = -2 + 4(-1) = -6$ $y' = 3 + 4(-1 - 3) = 3 + 4(-4) = -13$ New $L'(-6,-13)$.

3. For $M(2,-2)$: $x' = -2 + 4(2 - (-2)) = -2 + 4(4) = 14$ $y' = 3 + 4(-2 - 3) = 3 + 4(-5) = -17$ New $M'(14,-17)$.

Final coordinates: $K'(6,7), L'(-6,-13), M'(14,-17).$

Answer: C.

Question 12:

Find the area of triangle $ABC$ after dilation with center $O(0,0)$ and scale factor 4. Vertices of $ABC$: $A(-2,3)$, $B(2,3)$, $C(0,-4)$.

Original area: $\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$ Substitute: $\text{Area} = \frac{1}{2} \left| -2(3 - (-4)) + 2((-4) - 3) + 0(3 - 3) \right|$ $= \frac{1}{2} \left| -2(7) + 2(-7) \right| = \frac{1}{2} \left| -14 - 14 \right| = \frac{1}{2} \cdot 28 = 14$

Dilated area: Area scales by the square of the scale factor: $\text{Dilated area} = 4^2 \cdot 14 = 16 \cdot 14 = 224$

Answer: C.

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