Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | May-Jun | (P1-9709/13) | Q#8

Hits: 286



A(-1,1) and P(a,b) are two points, where a and b are constants. The gradient of AP is 2.

(i)        Find an expression for b in terms of a.

(ii)      B(10,-1) is a third point such that AP = AB. Calculate the coordinates of the possible positions           of P.



We are given coordinates of two points A(-1,1) and P(a,b).

We are also given gradient of line AP as;

Expression for slope of a line joining points  and ;

Therefore, for AP;


We are given that;

We will find both AB and AP and equate to find the coordinates of point P.

We are given coordinates of all points A(-1,1) and P(a,b) and B(10,-1).

As we have found in (i)  therefore coordinates of point P(a,2a+3)

Expression to find distance between two given points  and is:


Now we have two options.

We have found two possible x-coordinates of point P.

To find the corresponding possible y-coordinates of of point P, we can utilize the equation obtained  in (i);



Hence, possible coordinates of point P are;

Please follow and like us: