Question

Determine whether each set of points make a right triangle using the Pythagorean Theorem. 1. A(3 , -4) B(-4 , 3) C(0 , 0) 2. O(2 , 5) P(-1 , 3) Q(7 , 4) 3. T(1 , 1) U(3 , 3) V(5 , 1)

Answer 1

Answer:

$1.\ A(3 , -4)\ B(-4 , 3)\ C(0 , 0)$ -- Not right triangle

$2.\ O(2 , 5)\ P(-1 , 3)\ Q(7 , 4)$ -- Not right triangle

$3.\ T(1 , 1)\ U(3 , 3)\ V(5 , 1)$ -- Right triangle

Explanation:

Required

Determine whether the given points make a right triangle

$1.\ A(3 , -4)\ B(-4 , 3)\ C(0 , 0)$

First, calculate the distance between each point using:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

So, we have:

$AB = \sqrt{(3 - -4)^2 + (-4 - 3)^2}= \sqrt{98} = 7\sqrt 2$

$BC = \sqrt{(-4 - 0)^2 + (3 - 0)^2}= \sqrt{25} = 5$

$AC = \sqrt{(3 - 0)^2 + (-4 - 0)^2}= \sqrt{25} = 5$

From the above computations, the longest side is AB.

So;

$AB^2 = BC^2 + AC^2$ --- Test of Pythagoras

$(7\sqrt 2)^2 = 5^2 + 5^2$

$98 = 25 + 25$

$98 \ne 50$

The above points do not make a right triangle

$2.\ O(2 , 5)\ P(-1 , 3)\ Q(7 , 4)$

Calculate distance

$OP = \sqrt{(2 - -1)^2 + (5 - 3)^2}= \sqrt{13}$

$PQ = \sqrt{(-1 -7)^2 + (3 - 4)^2}= \sqrt{65}$

$OQ = \sqrt{(2 -7)^2 + (5 - 4)^2}= \sqrt{26}$

From the above computations, the longest side is PQ

So;

$PQ^2 = OP^2 + OQ^2$ --- Test of Pythagoras

$\sqrt{65}^2 = \sqrt{13}^2 + \sqrt{26}^2$

$65 = 13 + 26$

$65 \ne 39$

The above points do not make a right triangle

$3.\ T(1 , 1)\ U(3 , 3)\ V(5 , 1)$

Calculate distance

$TU = \sqrt{(1 -3)^2 + (1 - 3)^2}= \sqrt{8} = 2\sqrt2$

$UV = \sqrt{(3 -5)^2 + (1 - 3)^2}= \sqrt{8} = 2\sqrt2$

$TV = \sqrt{(1 -5)^2 + (1 - 1)^2}= \sqrt{16} = 4$

From the above computations, the longest side is TV

So;

$TV^2 = TU^2 + UV^2$ --- Test of Pythagoras

$4^2 = (2\sqrt 2)^2 +(2\sqrt 2)^2$

$16 = 8 + 8$

$16 = 16$

The above points not make a right triangle