# 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;