# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2006 | June | Q#11

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**Question**

The line passes through the points P(–1,2) and Q(11, 8).

**a. **Find an equation for in the form y= mx + c, where m and c are constants.

The line passes through the point R(10, 0) and is perpendicular to . The lines and intersect

at the point S.

**b. **Calculate the coordinates of S.

**c. **Show that the length of RS is .

**d. **Hence, or otherwise, find the exact area of triangle PQR.

**Solution**

**a.
**

We are given that passes through the points P(–1,2) and Q(11, 8).

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

Therefore, for line equation can be derived from coordinates of points P(–1,2) and Q(11, 8).

**b.
**

We are required to find the coordinates of point S which is intersection point of the lines and .

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).

We already have found equation of line in (a).

However, we need 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 are given that line passes through the point R(10, 0), hence, we have coordinates of a point on the line line.

We need slope of line to write its equation.

We are given that line is perpendicular to .

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

Therefore, if we have slope of line , we can find slope of line .

Slope-Intercept form of the equation of the line;

Where is the slope of the line.

We already have found equation of line in (a).

Therefore, slope of line is . Therefore;

Now, with coordinates of a point (10,0) on line and slope of line we can write equation of line .

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

To find the coordinates of point S which is intersection point of the lines and we need to solve the equations of these two lines.

Equating both equations;

Single value of x indicates that there is only one intersection point.

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;

Substitute x=7;

Hence, coordinates of point S(7,6).

**c.
**

We are required to find the length of RS.

Expression to find distance between two given points and is:

We are given coordinates of point R(10,0) and we have found in (b) coordinates of point S(7,6).

Therefore;

Since ;

**d.
**

Expression for the area of the triangle is;

We need both base and height of triangle PQR to find area.

If we sketch the points P, Q, R and S with given and found coordinates we get the following. Since we are given that line and are perpendicular, RS line makes the height of the triangle PQR and PQ make base of the same triangle.

We have already found in (c) that .

We need to find PQ.

Expression to find distance between two given points and is:

We are given coordinates of the points P(–1,2) and Q(11, 8).

Therefore;

Since ;

Hence;

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