O-Level Coordinate geometry in two dimensions: worked solution

Question

For triangle PQR with P(5, 4), R(9, 2), S(−1.5, 11), PR ∥ SQ and PQ: y = 2x − 6, find Q.

Q is the point where line SQ meets line PQ, so the plan is: find the equation of SQ, then solve it simultaneously with PQ. Step 1: find the gradient of PR, using P(5, 4) and R(9, 2) in gradient = (y₂ − y₁)/(x₂ − x₁). gradient of PR = (2 − 4)/(9 − 5) = −2/4 = −1/2 Step 2: PR is parallel to SQ, and parallel lines have equal gradients, so SQ also has gradient −1/2. Step 3: write the equation of SQ. It passes through S(−1.5, 11) with gradient −1/2, so use y − y₁ = m(x − x₁). y − 11 = −1/2 (x − (−1.5)) y − 11 = −0.5(x + 1.5) Expand the bracket: y − 11 = −0.5x − 0.75 Add 11 to both sides: y = −0.5x + 10.25 Step 4: Q lies on SQ and also on PQ, whose equation y = 2x − 6 is given. So the coordinates of Q satisfy both, and the two expressions for y can be set equal. 2x − 6 = −0.5x + 10.25 Step 5: solve for x. Add 0.5x to both sides and add 6 to both sides. 2x + 0.5x = 10.25 + 6 2.5x = 16.25 x = 16.25/2.5 = 6.5 Step 6: substitute back into the simpler equation y = 2x − 6 to find y. y = 2(6.5) − 6 = 13 − 6 = 7 So Q is the point (6.5, 7).