Sec 4 E-Math Coordinate Geometry notes

The distance between A(x₁, y₁) and B(x₂, y₂) is the hypotenuse of a right triangle with legs (x₂ − x₁) and (y₂ − y₁): distance = √((x₂ − x₁)² + (y₂ − y₁)²). The midpoint M splits the segment in half — average each coordinate: M = ((x₁ + x₂)/2, (y₁ + y₂)/2).

1. 1Distance A(1, 2) to B(7, 10)

   √((7 − 1)² + (10 − 2)²) = √(36 + 64) = √100 = 10.

2. 2Midpoint of AB

   ((1 + 7)/2, (2 + 10)/2) = (4, 6).

3. 3Watch the negatives

   For (−2, 3) to (4, −5): (4 − (−2))² = 6² — bracket the subtractions.

The gradient (steepness) of the line through (x₁, y₁) and (x₂, y₂) is m = (y₂ − y₁)/(x₂ − x₁). Subtract coordinates in the same order on top and bottom, and the sign looks after itself. A positive gradient climbs left to right; a negative gradient falls; a horizontal line has gradient 0 and a vertical line an undefined gradient.

• Through (1, 2) and (5, 14): (14 − 2)/(5 − 1) = 12/4 = 3.
• Through (−2, 1) and (4, −3): (−3 − 1)/(4 − (−2)) = −4/6 = −2/3.
• Parallel lines carry the same gradient: any line parallel to one of gradient 3 also has gradient 3.

Every non-vertical line writes as y = mx + c: gradient m, y-intercept c. To find the equation, first get m, then substitute any known point to solve for c. With two points, compute m from them, then use either point.

1. 1Gradient 3 through (2, 1)

   Start y = 3x + c and substitute: 1 = 3(2) + c → c = −5.

2. 2Write the equation

   y = 3x − 5.

3. 3Verify

   Check the other given point(s) satisfy it before moving on.

Equations sometimes arrive disguised: 2x + 3y = 6 must be rearranged to y = (6 − 2x)/3 = −(2/3)x + 2 before the gradient −2/3 and intercept 2 can be read. Rearrange every term — dividing one term only is a classic slip.

Coordinates turn geometry into arithmetic. Show a quadrilateral is a parallelogram by proving both pairs of opposite sides equal (distance formula) or parallel (equal gradients). Show a right angle with the converse of Pythagoras: compute the three side lengths and test whether the two shorter squares sum to the largest. Area of a coordinate triangle with a horizontal or vertical side is ½ × base × height, reading the lengths straight off the axes.

1. 1Right angle check

   A(1, 1), B(4, 1), C(4, 5): AB = 3, BC = 4, CA = √(9 + 16) = 5. Since 3² + 4² = 5², the angle at B is 90°.

2. 2Area

   Area = ½ × 3 × 4 = 6 square units.

3. 3Parallelogram check

   P(1,1), Q(5,1), R(7,4), S(3,4): PQ and SR are both horizontal with length 4; PS and QR both have gradient 3/2. Two parallel pairs → parallelogram.

Structure your geometric proofs: state the property being shown (parallel, equal, right-angled), show the computation, then conclude with the property named. Bare calculations without the concluding sentence lose the final mark.

Applied questions dress the toolkit in context: towns on a map grid, robotic arms on a factory floor, or sprinklers on a lawn. The mathematics never changes — convert each sentence into coordinates first, then run distance, midpoint, gradient or equation as usual. One grid square equals one unit unless a scale says otherwise.

• "How far apart" → distance formula.
• "Halfway between" → midpoint.
• "Does the path stay straight / is the wall parallel to" → compare gradients.
• "Is the corner square" → converse of Pythagoras on the three corner-to-corner lengths.

Keep units in the final sentence only: compute in grid units, convert at the end (10 units × 100 m). Converting inside the calculation invites arithmetic slips and mixes up the answer’s units.