Question

Determine which of the following sets of three points constitute the vertices of a right triangle: (a) 3 + 5i,2 +2i,5i; (b)2i,3 + 5i,4 + i; (c)6 +4i,7 + 5i, 8 +4i

Answer 1

Answer:

Option (c) is correct

Step-by-step explanation:

Case (a)

A = 3 + 5i = (3, 5)

B = 2 + 2i = (2, 2)

C = 5i = (0, 5)

Use the distance formula to find the distance between two points

$AB = \sqrt{(2-3)^{2}+(2-5)^{2}}=\sqrt{10}$

$BC = \sqrt{(0-2)^{2}+(5-2)^{2}}=\sqrt{13}$

$CA = \sqrt{(0-3)^{2}+(5-5)^{2}}=\sqrt{9}$

For the triangle to be right angles triangle

$BC^{2}=AB^{2}+CA^{2}$

Here, it is not valid, so these are not the points of a right angled triangle.

Case (b)

A = 2i = (0, 2)

B = 3 + 5i = (3, 5)

C = 4 + i = (4, 1)

Use the distance formula to find the distance between two points

$AB = \sqrt{(3-0)^{2}+(5-2)^{2}}=\sqrt{18}$

$BC = \sqrt{(4-3)^{2}+(1-5)^{2}}=\sqrt{17}$

$CA = \sqrt{(4-0)^{2}+(1-2)^{2}}=\sqrt{17}$

For the triangle to be right angles triangle

$AB^{2}=BC^{2}+CA^{2}$

Here, it is not valid, so these are not the points of a right angled triangle.

Case (c)

A = 6 + 4i = (6, 4)

B = 7 + 5i = (7, 5)

C = 8 + 4i = (8, 4)

Use the distance formula to find the distance between two points

$AB = \sqrt{(7-6)^{2}+(5-4)^{2}}=\sqrt{2}$

$BC = \sqrt{(8-7)^{2}+(4-5)^{2}}=\sqrt{2}$

$CA = \sqrt{(8-6)^{2}+(4-4)^{2}}=\sqrt{4}$

For the triangle to be right angles triangle

$CA^{2}=BC^{2}+AB^{2}$

Here, it is valid, so these are the points of a right angled triangle.