Question

∆ABC has A(-3, 6), B(2, 1), and C(9, 5) as its vertices. The length of side AB is units. The length of side BC is units. The length of side AC is units. °.

Answer 1

Answer:

AB = 7.07 units

BC = 8.06 units

AC = 12.04 units

Step-by-step explanation:

To find the length for each side of the triangle, apply the distance formula between each pair of vertices.

AB

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(2--3)^2 + (1-6)^2} \\d = \sqrt{(5)^2 + (-5)^2} \\d = \sqrt{25 + 25} \\d = \sqrt{50} \\d=7.07$

BC

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(9-2)^2 + (5-1)^2} \\d = \sqrt{(7)^2 + (4)^2} \\d = \sqrt{49 + 16} \\d = \sqrt{65} \\d=8.06$

AC

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(9--3)^2 + (5-6)^2} \\d = \sqrt{(12)^2 + (-1)^2} \\d = \sqrt{144 + 1} \\d = \sqrt{145} \\d=12.04$