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

`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

Step-by-step 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 = √((x_1 - x_2)^2 + (y_1 - y_2)^2)`

So, we have:

`AB = √((3 - -4)^2 + (-4 - 3)^2)= √(98) = 7\sqrt 2`

`BC = √((-4 - 0)^2 + (3 - 0)^2)= √(25) = 5`

`AC = √((3 - 0)^2 + (-4 - 0)^2)= √(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 \\e 50`

The above points do not make a right triangle

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

Calculate distance

`OP = √((2 - -1)^2 + (5 - 3)^2)= √(13)`

`PQ = √((-1 -7)^2 + (3 - 4)^2)= √(65)`

`OQ = √((2 -7)^2 + (5 - 4)^2)= √(26)`

From the above computations, the longest side is PQ

So;

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

`√(65)^2 = √(13)^2 + √(26)^2`

`65 = 13 + 26`

`65 \\e 39`

The above points do not make a right triangle

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

Calculate distance

`TU = √((1 -3)^2 + (1 - 3)^2)= √(8) = 2\sqrt2`

`UV = √((3 -5)^2 + (1 - 3)^2)= √(8) = 2\sqrt2`

`TV = √((1 -5)^2 + (1 - 1)^2)= √(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