OCR Advanced FSMQ - Additional Maths

Answer 12 (a) (i) 2 x + 5 y = 40 Hence 5 y = −2 x + 40 y = − 2 5 x + 8 Comparing this with the equation for a straight line, y = mx + c we obtain, gradient = − 2 5 (ii) The line parallel to 2 x + 5 y = 40 will have gradient − 2 5 . The equation of a line with gradient − 2 5 and passing through the point P (0, 6) is y − y 1 = m ( x − x 1 ) y − 6 = − 2 5 ( x − 0 ) 5 y − 30 = −2 x Hence, equation of line is 2 x + 5 y = 30 (b) The coordinates (5, p ) lie on the line, so these coordinates will satisfy the equation for the line. So, 5 y + 2 x = 30 5 p + 10 = 30 5 p = 20 p = 4 6.13 Proving that a point ( x , y ) lies on a line To prove that a point ( x , y ) lies on a line you substitute the coordinates of the point for x and y into the equation. If the left-hand side of the equation equals the right-hand side, the point lies on the line. Take the following equation of a straight line as an example: 2 y − 5 x = 7 To obtain the gradient, we need to rearrange this equation into the form y  = mx + c . The gradient will be given by the value of m . Parallel lines have the same gradient. > > >  TIP You could use the other method here where you let the equation of the parallel line be 2 x  + 5 y  =  c , and then substitute the coordinates of the points through which the line passes into the equation to find the value of c . Once this is done, this will be the required equation. x = 5 and y = p are substituted into the equation of the straight line and the resulting equation is solved to find the numerical value of p . 2 Coordinate geometry 134