Q6 Cambridge International Examinations

Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level May-June 2015 MATHEMATICS (US) Paper 1 Pure Mathematics 1 (P1) Q6 9280-11 Paper 1 Pure Mathematics 1 (P1) May-June 2015 shahbaz ahmed June 2024 1 Q6 The line with slope −2 passing through the point P (3t, 2t) intersects the x-axis at A and the y-axis at B. (i) Find the area of triangle AOB in terms of t. The line through P perpendicular to AB intersects the x-axis at C. (ii) Show that the mid-point of PC lies on the line y = x. Solution(i) Let m be the slope.Then m = −2 Equation of the line passing through the point P (3t, 2t) with the slope m = −2 2 y − 2t = −2(x − 3t)................ eq(1) Put y = 0 0 − 2t = −2(x − 3t) −2t = −2(x − 3t) t = x − 3t x − 3t = t x = 4t =⇒ A(4t, 0) is x-intercept. =⇒ |OA| = 4t Similarly putting x = 0 in the equation (1). y − 2t = −2(0 − 3t) y − 2t = 6t y = 8t B(0, 8t) is the y-intercept. =⇒ |OB| = 8t Area of the triangle AOB= 12 |OA||OB| Putting |OA| = 4t and |OB| = 8t Area of the triangle AOB= 12 (4t)(8t) = 16t2 (ii) Let m1 be the slope of the line through P perpendicular to AB .Then mm1 = −1 Putting m = −2 −2m1 = −1 m1 = 1 2 Equation of the line through the point P (3t, 2t) with slope m1 = y − 2t = 12 (x − 3t) 2y − 4t = x − 3t Putting y = 0 3 1 2 −4t = x − 3t x − 3t = −4t x = −t C(−t, 0) is the x-intercept of the line passing through P (3t, 2t) perpendicular to AB. The mid point of PC=( 3t+(−t) , 2t+0 2 2 ) = (t, t) Putting (x, y) = (t, t)in the equation. y=x t=t L.H.S=R.H.S So that the mid point (t, t) lies on the line y = x. 4