OCR Advanced FSMQ - Additional Maths

Multiplying through by 10 gives 10 y − 15 = −4( x − 2) 10 y − 15 = −4 x + 8 4 x + 10 y − 23 = 0 2 The points A , B , C , D have coordinates (−4, 4), (−1, 3), (0, 1), ( k , 0) respectively. The straight line CD is parallel to the straight line AB . (a) Find the gradient of AB . (b) Find the gradient of CD and hence find the value of the constant k . (c) Line L is perpendicular to CD and passes through point C . Find the equation of line L in the form ax + by + c = 0. Answer 2 Gradient of AB = y 2 − y 1 x 2 − x 1 = 3 − 4 −1 − (−4) = − 1 3 (b) Gradient of CD = y 2 − y 1 x 2 − x 1 = 0 − 1 k − 0 = − 1 k As line CD is parallel to AB the gradients are equal. Hence, − 1 3 = − 1 k Giving k = 3 This means D is the point (3, 0) and this point can be added to the graph. > > >  TIP There are lots of points in the question so spend the time drawing the points on a set of axes. Keep the scale on the x- and y -axes the same. As you proceed through the question you can add points and lines to your graph. D (3, 0) y x 0 –3 –2 –1 1 1 –4 2 3 2 3 4 A (–4, 4) B (–1, 3) C (0, 1) >> >> The fact that lines AB and CD are parallel can be shown by drawing double-headed arrows halfway along each line. Notice how you can use the graph to check the signs of the gradients of any lines. A quick look at the graph reveals the gradients of lines AB , BC and CD are all negative. The gradients are equated here. 6 Coordinate geometry of straight lines 139