1. **State the problem:** We have points X(-3,3), Y(3,1), Z(4,4), and W(2,-2). We need to find: - Distance from X to line WZ - Length XZ - Length XY - Distance from line $\ell$ (through W and Z) to point X Then identify which distances are the same and which is different, rounding to the nearest tenth. 2. **Find the distance from point X to line WZ:** The formula for distance from point $P(x_0,y_0)$ to line $Ax+By+C=0$ is: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ 3. **Find equation of line WZ:** Slope $m = \frac{4 - (-2)}{4 - 2} = \frac{6}{2} = 3$ Using point-slope form with W(2,-2): $$y - (-2) = 3(x - 2) \Rightarrow y + 2 = 3x - 6 \Rightarrow 3x - y - 8 = 0$$ So, $A=3, B=-1, C=-8$. 4. **Calculate distance from X(-3,3) to line WZ:** $$d = \frac{|3(-3) - 1(3) - 8|}{\sqrt{3^2 + (-1)^2}} = \frac{|-9 - 3 - 8|}{\sqrt{9 + 1}} = \frac{|-20|}{\sqrt{10}} = \frac{20}{\sqrt{10}}$$ Simplify: $$d = \frac{20}{\sqrt{10}} = \frac{20}{\cancel{\sqrt{10}}} \times \frac{\cancel{\sqrt{10}}}{\sqrt{10}} = \frac{20\sqrt{10}}{10} = 2\sqrt{10} \approx 6.3$$ 5. **Find length XZ:** Distance formula: $$XZ = \sqrt{(4 - (-3))^2 + (4 - 3)^2} = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \approx 7.1$$ 6. **Find length XY:** $$XY = \sqrt{(3 - (-3))^2 + (1 - 3)^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \approx 6.3$$ 7. **Distance from line $\ell$ (WZ) to point X is the same as step 4:** $$6.3$$ 8. **Summary:** - Distance from X to line WZ: $6.3$ - Length XY: $6.3$ - Length XZ: $7.1$ The "same" answer is about 6.3 units. The "different" answer is about 7.1 units.