# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2011 | June | Q#3

Question

The points P and Q have coordinates (–1, 6) and (9, 0) respectively.

The line is perpendicular to PQ and passes through the mid-point of PQ.

Find an equation for , giving your answer in the form ax + by + c = 0, where a, b and c are integers.

Solution

We are required to find equation of line .

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 have neither coordinates of a point on line nor slope of the line .

First we find coordinates of a point on line .

We are given the line passes through the mid-point of the line PQ and is also perpendicular to PQ.

We first find coordinates of the mid-point of line PQ.

We are given that points P and Q have coordinates (–1, 6) and (9, 0) respectively.

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 Therefore;

x-coordinate of mid-point of the PQ y-coordinate of mid-point of the PQ Hence, coordinates of mid-point of line PQ are (4,3).

Next, we find slope of line .

We are given the line is perpendicular to PQ.

If a line is normal to the curve , then product of their slopes and at that point (where line  is normal to the curve) is;   Therefore; Hence, if we have slope of the line PQ we can find slope of the line .

We can find slope of line PQ as follows.

Expression for slope (gradient) of a line joining points and ; We are given that points P and Q have coordinates (–1, 6) and (9, 0) respectively.

Therefore;    Hence, gradient of the line ;   With coordinates of a point M (4,3) on the line and slope of the line at hand , we can write equation of this line.

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