Triangle Type Parabola 70D03E

1. **Problem 1: Classify $\triangle JKL$ with vertices $J(-2,3)$, $K(1,6)$, and $L(5,-1)$.\n\n2. Calculate the lengths of sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.\n\n3. Length $JK$: $$\sqrt{(1 - (-2))^2 + (6 - 3)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}.$$\n\n4. Length $KL$: $$\sqrt{(5 - 1)^2 + (-1 - 6)^2} = \sqrt{4^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65}.$$\n\n5. Length $JL$: $$\sqrt{(5 - (-2))^2 + (-1 - 3)^2} = \sqrt{7^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}.$$\n\n6. Since $KL = JL = \sqrt{65}$, two sides are equal, so the triangle is isosceles.\n\n7. Check if the triangle is right by verifying the Pythagorean theorem: $JK^2 + KL^2 \stackrel{?}{=} JL^2$.\n\n8. Calculate squares: $JK^2 = 18$, $KL^2 = 65$, $JL^2 = 65$.\n\n9. $JK^2 + KL^2 = 18 + 65 = 83 \neq 65 = JL^2$, so not right angled.\n\n10. **Conclusion:** The triangle is non-right isosceles.\n\n---\n\n11. **Problem 2: Find the equation of a parabola for a satellite dish 24 feet wide and 10 feet tall, opening downward, vertex at $(0,10)$.\n\n12. The parabola is symmetric about the y-axis, so equation form: $$y = ax^2 + k,$$ with vertex $(0,k) = (0,10)$.\n\n13. Since it opens downward, $a < 0$.\n\n14. The dish width is 24 feet, so x-intercepts at $x = \pm 12$ where $y=0$.\n\n15. Use point $(12,0)$ to find $a$: $$0 = a(12)^2 + 10 \Rightarrow 0 = 144a + 10 \Rightarrow 144a = -10 \Rightarrow a = -\frac{10}{144} = -\frac{5}{72}.$$\n\n16. Equation is $$y = -\frac{5}{72}x^2 + 10.$$\n\n**Final answers:**\n- Triangle type: non-right isosceles.\n- Parabola equation: $y = -\frac{5}{72}x^2 + 10$.