Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2010  MayJun  (P19709/13)  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 yaxis. 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 (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope 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 midpoint of the line AC. Therefore, we can find coordinates of the point E as follows.
To find the midpoint of a line we must have the coordinates of the endpoints of the line.
Expressions for coordinates of midpoint of a line joining points and;
xcoordinate of midpoint of the line
ycoordinate of midpoint 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.
PointSlope form of the equation of the line is;
ii.
It is evident from the diagram in (i) that point B is the yintercept of the line EB.
The point at which curve (or line) intercepts yaxis, 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 midpoint of line BD.
To find the midpoint of a line we must have the coordinates of the endpoints of the line.
Expressions for coordinates of midpoint of a line joining points and;
xcoordinate of midpoint of the line
ycoordinate of midpoint 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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