# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | May-Jun | (P1-9709/13) | Q#6

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Question

The coordinates of points A and B are (3k – 1, k + 3) and (k + 3, 3k + 5)  respectively, where k is a constant (k 1).

i.       Find and simplify the gradient of AB, showing that it is independent of k.

ii.       Find and simplify the equation of the perpendicular bisector of AB.

Solution

i.

We are given coordinates of two points A(3k – 1, k + 3) and B(k + 3, 3k + 5) and  are required to find  gradient of AB.

Expression for slope of a line joining points and ; Therefore;        ii.

Next we are required to find the equation of the perpendicular bisector of AB.

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).

Therefore, to write equation of the perpendicular bisector of AB we need coordinates  of a point on the perpendicular bisector of AB and slope of the perpendicular bisector  of AB.

Let us first find coordinates of a point on the perpendicular bisector of AB.

Since it is the bisector of AB the mid-point of AB also lies on the perpendicular bisector of AB.

Therefore, we find coordinates of the mid-point of AB and same will be also the  coordinates of the point on the perpendicular bisector of AB.

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 We are given coordinates of two points A(3k – 1, k + 3) and B(k + 3, 3k + 5).

Hence;

x-coordinate of mid-point of the line y-coordinate of mid-point of the line x-coordinate of mid-point of the line y-coordinate of mid-point of the line Hence, coordinates of mid-point of AB are and these are also coordinates of a point on the perpendicular bisector of AB.

Next we need to find slope of the perpendicular bisector of AB.

If two lines are perpendicular (normal) to each other, then product of their slopes and is;  Therefore, if we can find slope of the AB, from this we can find slope of the  perpendicular bisector of AB.

As demonstrated in (i); Hence;   Now we can write equation of the perpendicular bisector of AB.

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