# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2005 | Oct-Nov | (P1-9709/01) | Q#7

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**Question**

Three points have coordinates A (2, 6), B (8, 10) and C (6, 0). The perpendicular bisector of AB meets the line BC at D. Find

** i. **the equation of the perpendicular bisector of AB in the form ax + by = c,

** ii. **the coordinates of D.

**Solution**

i.

To write the equation of the 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).

The point on perpendicular bisector of line AB is the mid-point of AB. So find out mid-point of AB first;

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

In the given case, end-points of the line AB are A (2, 6), B (8, 10);

x-coordinate of mid-point of the line AB;

y-coordinate of mid-point of the line AB;

Hence the point on perpendicular bisector of line AB is .

Now we need slope of perpendicular bisector of line AB to write its equation;

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

In this case;

Therefore, we need slope of the line AB to find slope of its perpendicular bisector;

Expression for slope of a line joining points and ;

Hence slope of the line AB with points A (2, 6), B (8, 10);

Using;

Now, with point and slope , we are ready to write the equation of the perpendicular bisector of the line AB;

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

ii.

Since D is the point of intersection of the perpendicular bisector of the line AB and line BC, we can find its coordinates by;

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 can equate coordinates of both lines i.e. equate equations of both the lines (or the line and the

curve).

In this case we already have the equation of the perpendicular bisector of the line AB from (i);

We can rewite the equation as;

We need to find equation of the line BC for which we have coordinates of points B (8, 10) and C (6,

0).

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

Therefore;

Now we have equations of both lines which intersect at point D; so we can equate the y-coordinate of both lines at point of intersection.

With x-coordinate of point of intersection of two lines (or line and the curve) at hand, we can find the y-coordinate of the point of intersection of two lines (or line and the curve) by substituting value of x-coordinate of the point of intersection in any of the two equations;

We choose the equation of line BC;

Hence the coordinates of point .

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