If A(3, 2), B(8, 3), and C(5, x) are the vertices of a right triangle with right angle C, find all possible values of x. (Enter your answers as a comma-separated list.)

Question

If A(3, 2), B(8, 3), and C(5, x) are the vertices of a right triangle with right angle C, find all possible values of x. (Enter your answers as a comma-separated list.)

Answer 1

There are three possible ways a right triangle can be made

1) AB is perpendicular to AC

2) AB is perpendicular to BC

3) AC is perpendicular to BC

1) Suppose AB is perpendicular to AC

slope AB = (3-2) /(8-3) = 1/5

Therefore the slope of AC has to be -5

slope of AC = (x - 2)/(5-3) = -5

(x - 2) / 2 = -5

x - 2 = -10

x = -8

3) Suppose AC is perpendicular to BC

slope AC = (x-2) /(5-3) = (x - 2)/2

slope of BC = (x - 3)/(5 - 8) = -(x - 3)/3

The slope of AC must be the negative reciprocal of the slope of BC

Therefore, (x-2) / 2 = 3 / (x-3)

Cross multiply: (x-2)(x-3) = 6

FOIL: x² - 5x + 6 = 6

x² - 5x = 0

x(x - 5) = 0

x = 0, 5

I'll let you try the last possibility on your own

Answer 2

The trick here that the other experts didn't catch is that angle C is the right angle, thus eliminating some of their possibilities. Point C will fall on the line x = 5 and thus there are only two solutions to this problem, one on each side of line AB, if C is the right angle. Let point C be given as (5, x).

Now, the length of line AC is sqrt((2 - x)² + (3 - 5)²) = sqrt((2 - x)² + 4).

The length of line BC is sqrt((3 - x)² + (8 - 5)²) = sqrt((3 - x)² + 9)).

The length of line AB is sqrt((2 - 3)² + (3 - 8)²) = sqrt(26).

Since C is the right angle, AB is the hypotenuse. By the Pythagorean Theorem, we have that AC² + BC² = AB². By substitution,

(2 - x)² + 4 + (3 - x)² + 9 = 26

4 - 4x + x² + 4 + 9 - 6x + x² +9 = 26

2x² -10x + 26 = 26

2x² - 10x = 0

x² - 5x = 0

x(x - 5) = 0

Thus, x = 0 or x = 5 and the solution set is C = {(5, 0), (5, 5)}.

Answer 3

(3,2), (8,3), (5,x)

AB = sqr((3-2)^2+ (8-3)^2) = sqr26

BC = sqr(9+(3-x)^2) = sqr(18-6x+x^2)

AC = sqr(4+(x-2)^2) = sqr(x^2-4x+8)

to be a right triangle one side squared = sum of the other two sides squared;

three possibilities:

AB^2 = BC^2 + AC^2

BC^2 = AB^2 +AC^2

AC^2 = AB^2 + BC^2

try the last one

26 +18-6x+x^2 = x^2-4x+8

2x =36

x=18

(5,18) is a 3rd point that makes the 3 points vertices of a right triangle

AC^2 = 260 = AB + BC = 26 + 4+ 225 = 260

make AB the hypotenuse, then BC the hypotenuse

set each = sum of the other two sides and solve for x

x=0,5,-8,

and x=18

(5,0), (5,5), (5-8) and (5,18) all create right triangles with points A and B

It may help to plot the points and connect them with points A and B, to visualize the right triangles.