# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2010 | May-Jun | (P1-9709/13) | Q#8

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Question

The diagram shows a rhombus ABCD in which the point A is (1, 2), the point C is (5, 4) and the point B lies on the y-axis. Find

i.       the equation of the perpendicular bisector of AC,

ii.       the coordinates of B and D,

iii.       the area of the rhombus.

Solution

i.

Consider the diagram below.

It is evident that perpendicular bisector of AC is the line EB.

To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).

The lines AC & EB intersect at point E.

If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e. coordinates of that point have same values on both lines (or on the line and the curve).

Therefore, we have point E on both the lines AC & EB. So we need coordinates of point E to write the equation of EB.

Since EB is perpendicular bisector of AC, point E is the mid-point of the line AC. Therefore, we can find coordinates of the point E as follows.

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

Hence;

Therefore, coordinates of point .

Now we need slope of the line EB to write its equation.

Since EB is perpendicular to AC, we can find its slope from slope of AC.

If two lines are perpendicular (normal) to each other, then product of their slopes  and  is;

Therefore, we find slope of AC;

Expression for slope (gradient) of a line joining points  and ;

Hence;

Now, we can write equation of the line EB.

Point-Slope form of the equation of the line is;

ii.

It is evident from the diagram in (i) that point B is the y-intercept of the line EB.

The point  at which curve (or line) intercepts y-axis, the value of . So we can find the value of  coordinate by substituting  in the equation of the curve (or line).

We have equation of EB from (i);

Hence coordinates of point .

Consider the diagram in (i) again.

It is evident that E is the mid-point of line BD.

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

From (i), we have coordinates of point  and from (ii) we have coordinates of point .

Therefore;

Hence coordinates of point .

iii.

Expression for the area of the rhombus is;

We can see from the diagram that two diagonals of given rhombus are  and .

Now we find  and .

Expression to find distance between two given points  and is:

We are given that coordinates of point A(1, 2) and point C (5, 4). Therefore;

Next
we find
. From (ii) we have
coordinates of point
and point . Therefore;

Expression to find distance between two given points  and is:

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