# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2007 | June | Q#11

Question

The line  has equation y=3x+2 and the line  has equation 3x+2y-8=0.

a.   Find the gradient of line for .

The point of intersection of  and  is P.

b.   Find the coordinates of P.

The lines  and  cross the line y=1 at the points A and B respectively.

c.   Find the area of triangle ABP.

Solution

a.

We are given equation of the line  as;

We are required to find the gradient of the line .

Slope-Intercept form of the equation of the line;

Where  is the slope of the line.

Therefore, we can find slope of the line  by rearranging its equation in terms slope-intercept form  as follows.

Therefore, slope of the line  is;

b.

We are required to find the coordinates of point P 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 are given equation of line ;

We are given equation of line ;

However, as demonstrated in (a) equation of line  can be written as;

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 ;

Hence, coordinates of point P.

c.

We are required to find the area of triangle ABP.

Expression for the area of the triangle is;

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

We need to locate the points A, B and P to sketch the triangle ABP.

For this we need coordinates of points A and B.

We are given that points A and B are intersection points of lines  and  with the line y=1,  respectively. Therefore, y-coordinate of both intersection points A and B will be same as y=1.

We need to find x-coordinates of both points A and B.

With y-coordinate of a point on the line or the curve at hand, we can find the x-coordinate of the  point by substituting value of y-coordinate of the point in equation of the line or the curve.

Since point A is intersection of lines  and y=1, therefore, it also lies on both lines   and y=1.

Hence, with y-coordinate of point A (y=1) at hand we can substitute it in equation of line   to find its  x-coordinate.

Therefore, coordinates of point .

Similarly, we find coordinates of point B.

Since point B is intersection of lines  and y=1, therefore, it also lies on both lines   and y=1.

Hence, with y-coordinate of point B (y=1) at hand we can substitute it in equation of line   to find its  x-coordinate.

Therefore, coordinates of point .

We locate points A, B and P and sketch the triangle ABP as shown below.

It is evident from the diagram that;

Base of Triangle ABP=AB

Height of Triangle ABP=PQ

We need to find these lights to calculate area of triangle ABP.

Expression to find distance between two given points  and is:

Therefore, we need coordinates of all points A, B, P and Q.

We have already found coordinates of points  and . We have also found coordinates  of point P in (b).

We only need to find coordinates of point Q. It is evident from the diagram that point Q lies in the  line y=1, therefore, y-coordinate of point Q is ‘1’.

It is also evident that PQ is perpendicular to line y=1, therefore, x-coordinate of point Q on line y=1 right under point P must be equal to x-coordinate of point Q ie .

Therefore, coordinates of point  .

 For Base of Triangle ABP=AB For Height of Triangle ABP=PQ

Now we can find area of triangle ABP.