Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2007  June  Q#11
Question
The line has equation y=3x+2 and the line has equation 3x+2y8=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 .
SlopeIntercept 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 slopeintercept 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 xcoordinate of point of intersection of two lines (or line and the curve) at hand, we can find the ycoordinate of the point of intersection of two lines (or line and the curve) by substituting value of xcoordinate 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, ycoordinate of both intersection points A and B will be same as y=1.
We need to find xcoordinates of both points A and B.
With ycoordinate of a point on the line or the curve at hand, we can find the xcoordinate of the point by substituting value of ycoordinate 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 ycoordinate of point A (y=1) at hand we can substitute it in equation of line to find its xcoordinate.
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 ycoordinate of point B (y=1) at hand we can substitute it in equation of line to find its xcoordinate.
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, ycoordinate of point Q is ‘1’.
It is also evident that PQ is perpendicular to line y=1, therefore, xcoordinate of point Q on line y=1 right under point P must be equal to xcoordinate 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.
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