Coordinate proofs
Placing figures on a coordinate grid lets algebra prove geometric facts. Distance, midpoint, and slope formulas do the work.
The three tools
The distance formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) compares lengths; the midpoint formula \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\) finds midpoints; and slope \(\frac{y_2 - y_1}{x_2 - x_1}\) tests parallel lines (equal slopes) and perpendicular lines (slopes multiplying to -1).
Classifying a triangle
Given vertices A(0, 0), B(4, 0), and C(2, 6), the lengths are AB = 4, AC = the square root of 40, and BC = the square root of 40, so the triangle is isosceles. Checking slopes shows whether any sides are perpendicular: AC has slope 3 and BC slope -3, whose product is -9, not -1, so there is no right angle.
Classifying a quadrilateral
For P(1, 1), Q(5, 2), R(6, 6), and S(2, 5): every side has length the square root of 17, so PQRS is a rhombus. Slopes of PQ and QR are 1/4 and 4, not negative reciprocals, so it is not a square. The diagonals PR and QS have slopes 1 and -1, so they are perpendicular, as in every rhombus.
Placing figures conveniently
In a general coordinate proof, place figures to make the algebra easy: put a vertex at the origin and a side along the x-axis. A rectangle can have vertices (0, 0), (a, 0), (a, b), and (0, b), so results hold for every rectangle, not just one example.
A general proof
To prove a rectangle's diagonals are congruent, use vertices (0, 0), (a, 0), (a, b), and (0, b). One diagonal runs from (0, 0) to (a, b), with length the square root of (a squared + b squared); the other runs from (a, 0) to (0, b), with the same length. Since a and b are arbitrary, the result holds for all rectangles.
Midsegment proof
Place a triangle at (0, 0), (2a, 0), and (2b, 2c). The midpoints of the two non-horizontal sides are (b, c) and (a + b, c), so the segment joining them is horizontal, parallel to the base, with length a, half the base length 2a. Using 2a, 2b, and 2c avoids fractions in the midpoints.
Partitioning a segment
To find the point two thirds of the way from A(1, 2) to B(10, 8), move two thirds of each change: \(x = 1 + \tfrac{2}{3}(9) = 7\) and \(y = 2 + \tfrac{2}{3}(6) = 6\), giving (7, 6). This divides AB in the ratio 2 : 1.
Worked example
A park designer wants a fountain equidistant from three gates at (0, 0), (8, 0), and (0, 6). The point equidistant from all three is the circumcentre; for a right triangle it is the midpoint of the hypotenuse, here the midpoint of (8, 0) and (0, 6), which is (4, 3). Each gate is 5 units away, which the distance formula confirms.
Key takeaways
Coordinate proofs use the distance formula for lengths, the midpoint formula for midpoints, and slopes for parallel and perpendicular lines. Classify triangles and quadrilaterals by comparing side lengths, slopes, and diagonals. Place figures conveniently, with a vertex at the origin and a side on an axis, and use variables so results hold generally, such as rectangle diagonals being congruent. Use 2a, 2b, 2c to avoid fractional midpoints. Partition segments by moving a fraction of each coordinate change. A right triangle's circumcentre is the midpoint of its hypotenuse.
Flashcard
Compares lengths
Check your understanding
Which point is two thirds of the way from A(1, 2) to B(10, 8)?