Coordinate geometry turns shapes into arithmetic. Once every point has an address (x, y), distances, midpoints, slopes and straight lines become formulas rather than drawings. Questions that look like they need a careful sketch - is this triangle isosceles, are these two lines perpendicular, where do they meet - are answered by substituting coordinates.
It is also the bridge back to algebra: a line on the grid is exactly a linear equation, so slope and intercept carry their usual meanings.
The model omits vertical lines, which have undefined slope and cannot be written as y = mx + c; they take the form x = a constant. That single exception is a favourite trap.
Mini-example: (1, 2) to (4, 6). Gaps are 4 − 1 = 3 and 6 − 2 = 4, so distance = √(9 + 16) = √25 = 5. The distance formula is Pythagoras with the horizontal and vertical gaps as the two short sides.
Mini-example: (1, 2) and (4, 6) again. Midpoint = ( (1 + 4) ÷ 2 , (2 + 6) ÷ 2 ) = (2.5, 4). Each coordinate is the average of the two endpoints — never a difference.
Mini-example: (1, 2) to (4, 6). Rise is 6 − 2 = 4, run is 4 − 1 = 3, so m = 4/3. If the line falls left to right, m is negative — the sign is a free check.
Mini-example with m = 4/3 through (1, 2): y − 2 = (4/3)(x − 1), so 3y − 6 = 4x − 4, giving 3y = 4x + 2 or y = (4/3)x + 2/3; the intercept c is 2/3. Use y = mx + c when you know the slope and want the intercept; use the point-slope form when you know the slope and one point. The y-intercept is c (set x = 0); the x-intercept is found by setting y = 0 and solving for x.
| Relationship | Slope condition | Example |
|---|---|---|
| Parallel | m1 = m2 | y = 2x + 1 is parallel to y = 2x − 7 |
| Perpendicular | m1 × m2 = −1 | y = 2x has a perpendicular slope of −1/2 |
A horizontal line has slope 0; a vertical line has undefined slope, so it has no perpendicular condition in this form.
Each formula earns a thirty-second check before it is used. Distance: gaps first, then squares, then the root — if your answer is larger than both gaps by a lot, you probably added instead of taking the root. Midpoint: average, never difference; the result must sit halfway, so each coordinate of the midpoint lies between the two endpoints. Slope: rise over run in the same order top-to-bottom as right-to-left; the sign of m must match the visual direction of the line. Line equation: expand before simplifying, then verify by substituting the second point — one substitution catches almost every algebra slip.
The points A(2, 1) and B(6, 9) are given. Find the distance AB, the midpoint, the slope, and the equation of the line through them.
Distance: the gaps are 6 − 2 = 4 and 9 − 1 = 8, so
Midpoint: average the coordinates, giving ( (2 + 6) ÷ 2 , (1 + 9) ÷ 2 ) = (4, 5).
Slope: m = (9 − 1) ÷ (6 − 2) = 8 ÷ 4 = 2.
Equation: use point-slope with A(2, 1) and m = 2. Then y − 1 = 2(x − 2), which expands to y = 2x − 4 + 1, so y = 2x − 3.
Check with B(6, 9): 2 × 6 − 3 = 9, which matches. The x-intercept is where y = 0, giving 2x = 3, so x = 1.5; a line perpendicular to this one would have slope −1/2.
What is the distance between (0, 0) and (6, 8)?
What is the slope of the line through (1, 2) and (4, 11)?
What is the midpoint of (−2, 5) and (6, −1)?
A line has slope 3. What is the slope of any line perpendicular to it?
Without looking back: write the slope formula and the condition that makes two lines perpendicular.
The slope is (y2 − y1) ÷ (x2 − x1); two lines are perpendicular when the product of their slopes is −1.
The slope of the line through two points is the change in y divided by the . For perpendicular lines, the two slopes are negative . The midpoint of (2, 4) and (8, 10) has x-coordinate and y-coordinate .
A triangle has vertices A(0, 0), B(4, 0) and C(0, 3). Find the length of each side, and decide whether the triangle is right-angled.
AB runs along the x-axis from 0 to 4, so AB = 4. AC runs along the y-axis from 0 to 3, so AC = 3.
BC uses the distance formula: √((4 − 0)² + (0 − 3)²) = √(16 + 9) = √25 = 5.
The sides are 3, 4 and 5. Since 3² + 4² = 9 + 16 = 25 = 5², the converse of Pythagoras confirms a right angle, and because 5 is the longest side it must be opposite the right angle - at vertex A.
That also follows from the slopes: AB is horizontal (slope 0) and AC is vertical (undefined slope), and the axes meet at a right angle. The area is (1/2) × 4 × 3 = 6 square units.
Coordinates joined algebra to geometry. The next step is to read information that arrives already organised - tables, charts and averages - and to turn that data back into quantitative conclusions.