Now that we know how to prove triangles are congruent, we can also do so when the triangles are given on the coordinate plane. In fact, having the triangles on the coordinate plane makes them drawn to scale.
Here are some things we need to review before we start coordinate proofs:
Distance formula The distance between \((x_1, y_1)\) and \((x_2, y_2)\) can be found using \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\). Need a review? Click Here
Midpoint formula: The midpoint between \((x_1, y_1)\) and \((x_2, y_2)\) can be found using \(\left(\Large\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\). Need a review? Click Here
Slope: To find the slope between \((x_1, y_1)\) and \((x_2, y_2)\) can be found using \(\Large\frac{y_1 - y_2}{x_1 - x_2}\). Need a review? Click Here
Parallel and perpendicular slopes: The slopes of parallel lines must be the same, where the slopes of perpendicular lines must be opposite reciprocals of each other. Need a review? Click Here
A coordinate proof is a proof that shows why we know segments are congruent or angles are congruent using the distance formula, midpoint formula, slope, and possibly other geometric reasoning that we have learned. Then we will still use the same postulates and theorems to prove the triangles are congruent. Let’s look at the example below.
Example 1:
Prove, using coordinate geometry, that \(\triangle ABC\cong\triangle DEF\).
Here are some things we need to review before we start coordinate proofs:
Distance formula The distance between \((x_1, y_1)\) and \((x_2, y_2)\) can be found using \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\). Need a review? Click Here
Midpoint formula: The midpoint between \((x_1, y_1)\) and \((x_2, y_2)\) can be found using \(\left(\Large\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\). Need a review? Click Here
Slope: To find the slope between \((x_1, y_1)\) and \((x_2, y_2)\) can be found using \(\Large\frac{y_1 - y_2}{x_1 - x_2}\). Need a review? Click Here
Parallel and perpendicular slopes: The slopes of parallel lines must be the same, where the slopes of perpendicular lines must be opposite reciprocals of each other. Need a review? Click Here
A coordinate proof is a proof that shows why we know segments are congruent or angles are congruent using the distance formula, midpoint formula, slope, and possibly other geometric reasoning that we have learned. Then we will still use the same postulates and theorems to prove the triangles are congruent. Let’s look at the example below.
Example 1:
Prove, using coordinate geometry, that \(\triangle ABC\cong\triangle DEF\).
Solution:
By looking at the diagram, it seems as though the triangles are a \(90^{\circ}\) rotation of each other. We are now trying to prove that this is an isometry!
We can use a couple of methods to prove that the triangles are congruent. It seems as though \(\angle D\) and \(\angle A\) are right angles. If we can show that two sides of each triangle have opposite reciprocal slopes, then this would prove that the sides are perpendicular and that \(\angle D\) and \(\angle A\) are right angles.
Let's check the slopes:
Slope of \(\overline{AB} = \Large\frac{6 - 2}{1 - (-2)} = \frac{4}{3}\)
Slope of \(\overline{AC} = \Large\frac{2 - 0}{-2 - 1} = \frac{2}{-3} = -\frac{2}{3}\)
Since \(\Large\frac{4}{3}\) and \(-\Large\frac{2}{3}\) are NOT opposite reciprocals, \(\overline{AB}\not\perp\overline{AC}\), so \(\angle A\) is not a right angle.
These triangles are not right triangles, so now we can only use side lengths to prove the triangles congruent. To find the lengths of sides, we must use the distance formula or the Pythagorean theorem:
For \(\triangle ABC\)
\(AB = \sqrt{(1 - -2)^2 + (6 - 2)^2} = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\) units
\(AC = \sqrt{(-2 - 1)^2 + (2 - 0)^2} = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13}\) units
From the graph \(BC = 6\) units
For \(\triangle DEF\)
\(DE = \sqrt{(-6 - -10)^2 + (-4 - -1)^2} = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\) units
\(DF = \sqrt{(-6 - -4)^2 + (-4 - -1)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}\) units
From the graph \(EF = 6\) units
By looking at the diagram, it seems as though the triangles are a \(90^{\circ}\) rotation of each other. We are now trying to prove that this is an isometry!
We can use a couple of methods to prove that the triangles are congruent. It seems as though \(\angle D\) and \(\angle A\) are right angles. If we can show that two sides of each triangle have opposite reciprocal slopes, then this would prove that the sides are perpendicular and that \(\angle D\) and \(\angle A\) are right angles.
Let's check the slopes:
Slope of \(\overline{AB} = \Large\frac{6 - 2}{1 - (-2)} = \frac{4}{3}\)
Slope of \(\overline{AC} = \Large\frac{2 - 0}{-2 - 1} = \frac{2}{-3} = -\frac{2}{3}\)
Since \(\Large\frac{4}{3}\) and \(-\Large\frac{2}{3}\) are NOT opposite reciprocals, \(\overline{AB}\not\perp\overline{AC}\), so \(\angle A\) is not a right angle.
These triangles are not right triangles, so now we can only use side lengths to prove the triangles congruent. To find the lengths of sides, we must use the distance formula or the Pythagorean theorem:
For \(\triangle ABC\)
\(AB = \sqrt{(1 - -2)^2 + (6 - 2)^2} = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\) units
\(AC = \sqrt{(-2 - 1)^2 + (2 - 0)^2} = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13}\) units
From the graph \(BC = 6\) units
For \(\triangle DEF\)
\(DE = \sqrt{(-6 - -10)^2 + (-4 - -1)^2} = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\) units
\(DF = \sqrt{(-6 - -4)^2 + (-4 - -1)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}\) units
From the graph \(EF = 6\) units
Notice, \(AB = DE = 5\) units. Therefore, \(\overline{AB}\cong\overline{DE}\). Since \(AC = DF = \sqrt{13}\) units, \(\overline{AC}\cong\overline{DF}\). Similarly, \(\overline{BC}\cong\overline{EF}\) because their lengths are both \(6\) units. Therefore, since we know three pairs of corresponding sides are congruent between two triangles, \(\triangle ABC\cong\triangle DEF\) by SSS.
Example 2:
Use coordinate geometry to prove or disprove that the triangles are congruent.
Example 2:
Use coordinate geometry to prove or disprove that the triangles are congruent.
Solution:
Let’s watch the video for the solution!
Let’s watch the video for the solution!
Quick Check
1) Prove using coordinate geometry that \(\triangle ABC\cong\triangle DEF\).
1) Prove using coordinate geometry that \(\triangle ABC\cong\triangle DEF\).
2) Prove using coordinate geometry that \(\triangle ABC\cong\triangle DBC\).