Question

determine whether a triangle with the given vertices is a right triangle.
(a) $p(6, 4), q(9, 10), r(-2, 8)$
(b) $d(4, 0), e(-1, 6), f(-4, -5)$
(c) $j(-7, 2), k(-3, 10), l(0, -2)$
options for each:
right triangle
not a right triangle
cannot be determined

Explanation:

To determine if a triangle is right-angled, we use the distance formula to find side lengths, then check if the Pythagorean theorem holds ($a^2 + b^2 = c^2$, where $c$ is the longest side).
Distance formula: For points $(x_1,y_1)$ and $(x_2,y_2)$, $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, so $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2$.

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(a) For $P(6, 4), Q(9, 10), R(-2, 8)$

Step1: Calculate $PQ^2$

$PQ^2 = (9-6)^2 + (10-4)^2 = 3^2 + 6^2 = 9 + 36 = 45$

Step2: Calculate $QR^2$

$QR^2 = (-2-9)^2 + (8-10)^2 = (-11)^2 + (-2)^2 = 121 + 4 = 125$

Step3: Calculate $RP^2$

$RP^2 = (6-(-2))^2 + (4-8)^2 = 8^2 + (-4)^2 = 64 + 16 = 80$

Step4: Check Pythagorean theorem

$45 + 80 = 125$, so $PQ^2 + RP^2 = QR^2$.

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(b) For $D(4, 0), E(-1, 6), F(-4, -5)$

Step1: Calculate $DE^2$

$DE^2 = (-1-4)^2 + (6-0)^2 = (-5)^2 + 6^2 = 25 + 36 = 61$

Step2: Calculate $EF^2$

$EF^2 = (-4-(-1))^2 + (-5-6)^2 = (-3)^2 + (-11)^2 = 9 + 121 = 130$

Step3: Calculate $FD^2$

$FD^2 = (4-(-4))^2 + (0-(-5))^2 = 8^2 + 5^2 = 64 + 25 = 89$

Step4: Check Pythagorean theorem

$61 + 89 = 150
eq 130$, $61 + 130 = 191
eq 89$, $89 + 130 = 219
eq 61$.

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(c) For $J(-7, 2), K(-3, 10), L(0, -2)$

Step1: Calculate $JK^2$

$JK^2 = (-3-(-7))^2 + (10-2)^2 = 4^2 + 8^2 = 16 + 64 = 80$

Step2: Calculate $KL^2$

$KL^2 = (0-(-3))^2 + (-2-10)^2 = 3^2 + (-12)^2 = 9 + 144 = 153$

Step3: Calculate $LJ^2$

$LJ^2 = (-7-0)^2 + (2-(-2))^2 = (-7)^2 + 4^2 = 49 + 16 = 65$

Step4: Check Pythagorean theorem

$80 + 65 = 145
eq 153$, $80 + 153 = 233
eq 65$, $65 + 153 = 218
eq 80$.

Answer:

(a) Right triangle
(b) Not a right triangle
(c) Not a right triangle