# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | May-Jun | (P1-9709/12) | Q#8

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Question

Three points have coordinates A(0,7), B(8,3) and C(3k,k). Find the value of the constant k for which

i.       C lies on the line that passes through A and B,

ii.       C lies on the perpendicular bisector of AB.

Solution

i.

If point C lies on line AB, then coordinates of point C must satisfy equation of line AB.

Let’s first find equation of line 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).

We are given coordinates of points A(0,7) and B(8,3).

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

Therefore for line AB;

Substitute coordinates of point C(3k,k);

ii.

If point C lies on perpendicular bisector of line AB, then coordinates of point C must satisfy equation  perpendicular bisector of line AB.

Let’s first find equation of perpendicular bisector of line 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).

It is evident that mid-point of line AB also lies on perpendicular bisector of line AB.

Let’s find coordinates of mid-point of line 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 points A(0,7) and B(8,3).

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Hence coordinates of mid-point of line AB are (4,5).

Now we find slope of perpendicular bisector of line AB.

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

Therefore;

Therefore, if we have slope of line AB, we can find slope of perpendicular bisector of line AB.

Expression for slope of a line joining points  and ;

We are given coordinates of points A(0,7) and B(8,3).

Therefore, slope of line AB;

Now we can find slope of perpendicular bisector of line AB.

With coordinates of a point (4,5) on perpendicular bisector and its slope , we can write  equation of perpendicular bisector of line AB.

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

Therefore;

Substituting coordinates of point C(3k,k);