# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2009 | Oct-Nov | (P1-9709/12) | Q#9

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Question The diagram shows a rectangle . The point is and is . The diagonal is parallel to the x-axis.

i.       Explain why the y-coordinate of is 6. The x-coordinate of is .

ii.       Express the gradients of and in terms of .

iii.       Calculate the x-coordinates of and .

iv.       Calculate the area of the rectangle .

Solution

i.

Consider the diagram below. We know that the two diagonals of the rectangle intersect at mid-point. Therefore, point E is the mid-point of both AC & BD, diagonals.

To find the mid-point of a line we must have the coordinates of the end-points of the line.

Expressions for coordinates of mid-point of a line joining points and ;

x-coordinate of mid-point of the line y-coordinate of mid-point of the line Therefore;

y-coordinate of mid-point  y-coordinate of mid-point  Since, E & D lay on same horizontal line i.e. BD parallel to x-axis. Their y-coordinates are same. ii.

From (i), we have; We are also given that Therefore; Expression for slope (gradient) of a line joining points and ;   iii.

From the given diagram, we can find the length of diagonal AC.

Expression to find distance between two given points and is: Therefore;   We know that diagonals of a rectangle are equal in length. Hence; We have the y-coordinate of the mid-point of the diagonal BD, from (i);

y-coordinate of mid-point  We can also find x-coordinate of the mid-point of the diagonal BD;

x-coordinate of mid-point  x-coordinate of mid-point  Therefore; Since; We have 10 units length on each side of mid-point E on the diagonal BD. Therefore;

x-coordinate of D x-coordinate of B Therefore;  iv.

Expression for the area of the rectangle is; For the given case;  Expression to find distance between two given points and is: Therefore;          Hence;   