Verify a Right Triangle on the Coordinate Plane | Grade 8 Math | LevelKite

A right angle measures exactly $90^\circ$ , and a right triangle is a triangle with one right angle. The two sides that form the right angle are the legs. The side across from the right angle is the hypotenuse, and it is always the longest side.

On a coordinate plane, a corner can look like a right angle while being off by a degree or two, a difference your eye can't reliably see. To verify that a triangle is right, you prove it from exact information: the coordinates of its vertices, or corner points. All lengths in this lesson are measured in grid units.

You already use the Pythagorean theorem: in a right triangle with legs of lengths $a$ and $b$ and hypotenuse of length $c$ , $a^2 + b^2 = c^2$ . That theorem starts from a triangle you know is right. To verify, you need to go the other way.

Picture two sticks, of lengths $a$ and $b$ , joined at one end by a hinge. As the hinge opens wider, the distance between the free ends grows, and so does its square. That square equals $a^2 + b^2$ at exactly one opening: a right angle.

This is the idea behind the converse of the Pythagorean theorem, which switches the theorem's if and then parts. If a triangle has positive side lengths $a$ , $b$ and $c$ , with $c$ the longest, and $a^2 + b^2 = c^2$ , then the triangle is right, and its right angle is across from side $c$ . If $a^2 + b^2 \neq c^2$ , the triangle is not right: a right angle could only be across from the longest side, because the hypotenuse is always longest, and there the Pythagorean theorem would make the equation true.

Key concept

Verifying a right triangle from coordinates

For a triangle whose vertices have exact coordinates:

Find the squared length of each side, the square of its length: $(\text{change in } x)^2 + (\text{change in } y)^2$ . It is your distance method, stopped before the square root.
Call the largest squared length $c^2$ . Lengths are positive, so the largest square belongs to the longest side.
Add the other two squared lengths. If the sum equals $c^2$ exactly, the triangle is right, with its right angle at the vertex across from the longest side. If not, the triangle is not right.

You never take a square root, so nothing gets rounded: with whole-number coordinates, every squared length is an exact whole number.

a^2 + b^2 \stackrel{?}{=} c^2 \quad (c \text{ is the longest side})

Point A is at

(1, 6)

, Point B is at

(3, 2)

and Point C is at

(5, 3)

; the corner at B looks like a right angle.

Worked example

The figure above shows triangle $ABC$ with vertices $A(1, 6)$ , $B(3, 2)$ and $C(5, 3)$ . The corner at $B$ looks like a right angle. Use the coordinates to decide whether triangle $ABC$ is a right triangle.

Find the squared length of side AB. Subtract the coordinates of A from those of B: the change in x is 2 and the change in y is −4. Square each change and add. Squaring turns −4 into 16, so it does not matter which point you subtract from which.
$AB^2 = (3 - 1)^2 + (2 - 6)^2 = 2^2 + (-4)^2 = 4 + 16 = 20$
Find the squared length of side BC the same way, using the changes from B to C.
$BC^2 = (5 - 3)^2 + (3 - 2)^2 = 2^2 + 1^2 = 4 + 1 = 5$
Find the squared length of side AC, using the changes from A to C.
$AC^2 = (5 - 1)^2 + (3 - 6)^2 = 4^2 + (-3)^2 = 16 + 9 = 25$
Compare the squared lengths 20, 5 and 25. The largest is 25. Lengths are positive, so the largest square comes from the longest side: AC is the longest side, and its squared length plays the role of c squared.
$c^2 = AC^2 = 25$
Add the two smaller squared lengths and compare the sum with 25. They match exactly, so by the converse of the Pythagorean theorem, triangle ABC is right. Its right angle is across from AC, which is the angle at vertex B.
$AB^2 + BC^2 = 20 + 5 = 25 = AC^2$

Answer

Yes. Since $AB^2 + BC^2 = 25 = AC^2$ and $AC$ is the longest side, triangle $ABC$ is a right triangle with its right angle at vertex $B$ .

Point D is at

(-3, 3)

, Point E is at

(-2, -2)

and Point F is at

(4, -1)

; the corner at E looks like a right angle.

Worked example

The figure above shows triangle $DEF$ with vertices $D(-3, 3)$ , $E(-2, -2)$ and $F(4, -1)$ . Its corner at $E$ also looks like a right angle. Is triangle $DEF$ a right triangle?

Find the squared length of DE. For the change in x, subtracting a negative number means adding its opposite: −2 − (−3) = −2 + 3 = 1. The change in y is −2 − 3 = −5.
$DE^2 = (-2 - (-3))^2 + (-2 - 3)^2 = 1^2 + (-5)^2 = 1 + 25 = 26$
Find the squared length of EF. Both subtractions take away a negative number, so both become additions: 4 + 2 = 6 and −1 + 2 = 1.
$EF^2 = (4 - (-2))^2 + (-1 - (-2))^2 = 6^2 + 1^2 = 36 + 1 = 37$
Find the squared length of DF. The change in x is 4 − (−3) = 4 + 3 = 7, and the change in y is −1 − 3 = −4.
$DF^2 = (4 - (-3))^2 + (-1 - 3)^2 = 7^2 + (-4)^2 = 49 + 16 = 65$
The largest squared length is 65, so DF is the longest side. Add the other two squared lengths and compare the sum with 65.
$DE^2 + EF^2 = 26 + 37 = 63 \neq 65 = DF^2$
The sum misses by 2, and close is not equal. If the angle at E were a right angle, the Pythagorean theorem would make the sum exactly 65, so the angle at E is not a right angle. The angles at D and F cannot be right angles either, because a right angle is always across from the longest side.

Answer

No. Since $DE^2 + EF^2 = 63$ but $DF^2 = 65$ , triangle $DEF$ is not a right triangle, even though its corner at $E$ looks like a right angle.

Worked example

Triangle $PQR$ has vertices $P(-1, 4)$ , $Q(4, -1)$ and $R(1, -2)$ , with no figure given. Is triangle $PQR$ a right triangle? If it is, at which vertex is its right angle?

You need all three squared lengths before you can tell which side is longest. Start with PQ: square the change in x and the change in y, then add.
$PQ^2 = (4 - (-1))^2 + (-1 - 4)^2 = 5^2 + (-5)^2 = 25 + 25 = 50$
Next, find the squared length of QR the same way.
$QR^2 = (1 - 4)^2 + (-2 - (-1))^2 = (-3)^2 + (-1)^2 = 9 + 1 = 10$
Last, find the squared length of RP.
$RP^2 = (-1 - 1)^2 + (4 - (-2))^2 = (-2)^2 + 6^2 = 4 + 36 = 40$
The largest squared length is 50, so PQ is the longest side. The other two add to exactly 50, so by the converse the triangle is right. Its right angle is across from PQ, at vertex R.
$QR^2 + RP^2 = 10 + 40 = 50 = PQ^2$

Answer

Yes. Triangle $PQR$ is a right triangle, and its right angle is at vertex $R$ , across from the longest side $PQ$ .

To verify that a triangle on the coordinate plane is right, trust the coordinates, not the picture: triangles $ABC$ and $DEF$ both looked right, but only $ABC$ is. Find the three squared lengths and use the largest as $c^2$ ; because lengths are positive, it belongs to the longest side. If the other two add to exactly $c^2$ , the triangle has a right angle across from the longest side. Any other result, even a near miss, means the triangle has no right angle.