# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2008 | January | Q#1

**Question**

The triangle ABC has vertices A(-2,3), B(4,1) and C(2,-5).

**a. **Find the coordinates of the mid-point of BC .

**b. **

** **

**i.**F ind the gradient of AB, in its simplest form.

** **

**ii.**Hence find an equation of the line AB , giving your answer in the form , where q and r are integers.

** ****iii. **F ind an equation of the line passing through C which is parallel to AB.

**c. **Prove that angle ABC is a right angle.

**Solution**

**a.
**

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

For the given case we have B(4,1) and C(2,-5).

x-coordinate of mid-point of the line

y-coordinate of mid-point of the line

Hence coordinates of mid-point of BC are (3,-2).

**b.
**

** i.
**

Expression for slope (gradient) of a line joining points and ;

For the given case we have A(-2,3) and B(4,1).

Hence gradient (slope) of AB is .

ii.

We are required to find equation of the 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 already have coordinates of both points A(-2,3) and B(4,1).

We also have found slope of line AB, in (b:i);

So we can use both two point and point-slope forms to write equation of the line AB.

We choose point-slope form with point B(4,1) and .

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

Therefore;

iii.

We are required to find equation of the line which is parallel to AB and passes through point C(2,-5).

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 already have coordinate of one point on the line C(2,-5).

We need slope of line which is parallel to AB.

If two lines are parallel to each other, then their slopes and are equal;

Therefore

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

Hence;

**c.
**

If angle ABC is right angle then lines AB and BC must be perpendicular.

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

We already have found slope of line AB in (b:i).

Now we find slope of line BC.

Expression for slope (gradient) of a line joining points and ;

For the given case we have B(4,1) and C(2,-5).

Hence gradient (slope) of BC is .

Therefore, lines AB and BC are perpendicular and hence angle ABC is right angle.

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