The coordinates of the vertices of ABC are A(-4,5) , B(-7,6) and C(-4,15). Algebraically determine whether or not ABC is a right triangle

Answer 1

Greetings! Let us solve this problem with the Distance Formula! We will find the distance to all three sides to determine if it is a right triangle. So, let

The Distance between A and B = d_AB

The Distance between B and C = d_BC

The Distance between A and C = d_AC

Now, we have

d_AB = √(-4-(-7))² + (5-6)² = √(3)² + (-1)² = √10

d_BC = √(-4-(-7))² + (15-6)² = √(3)²+(9)² = √90 = √(9*10) = 3√10

d_AC = √(-4-(-4))²+(15-5)² = √(0) + (10)² = √(100) = 10

Since √10 is the shortest side, 3√10 is the second leg and 10 is the longest leg, numerically, we use the Pythagorean Theorem to prove it as a right triangle by showing,

c² = a² + b²

Then,

(10)² = (√10)² + (3√10)²

100 = 10 + (3)²*(10), because the square root of (10)² is 10.

100 = 10 + 90

Therefore, ABC is indeed a right triangle.

Hope this helps! ◊

Answer 2

All you need to do is plot the three points and determined the slope of each of the lines. If any of the lines are perpendicular then their slopes will be negative reciprocals