Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2017  June  Q#8
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Question
The straight line , shown in Figure, has equation 5y = 4x + 10.
The point P with x coordinate 5 lies on .
The straight line is perpendicular to and passes through P.
a. Find an equation for , writing your answer in the form ax + by + c = 0 where a, b and c are integers.
The lines and cut the xaxis at the points S and T respectively, as shown in Figure.
b. Calculate the area of triangle SPT.
Solution
a.
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 (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope form of Equation of Line).
We have xcoordinates of a point P(5,y) on line but not slope of the line .
First we find ycoordinate of the point P on the line . Since this point also lies on the line , we can utilize equation of the line to find ycoordinate of point P(5,y).
If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve (or the line).
We are given that point P(5,y) lies on the line which has equation; therefore, we substitute x=5 in equation of the line ;
Hence, coordinates of point P are (5,6).
Next we need slope of the line to write its equation.
We are given that the 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 the line .
We need to find the gradient of .
SlopeIntercept form of the equation of the line;
Where is the slope of the line.
Therefore, we can rearrange the given equation of line in slopeintercept form, as follows, to find the gradient of the line.
Hence, gradient of the line is;
Therefore;
With coordinates of a point on the line as P(5,6) and its slope at hand, we can write equation of the line .
PointSlope form of the equation of the line is;
b.
We are required to find area of triangle SPT.
Expression for the area of the triangle with base and height is;
It is evident that triangle SPT is a right angled triangle with .
It is evident that PT can be considered as base of triangle SPT whereas SP is height of the triangle.
We need to find distances PT and SP.
Expression for the distance between two given points and is:
We do not have coordinates of the points S and T so let us find them first.
We have coordinates of P(5,6).
We need to find coordinates of point S, next.
Point S is the xintercept of the line .
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
We substitute y=0 in equation of the line ;
Hence, coordinates of xintercept of the line (point S) are .
Next we need coordinates of point T which is the xintercept of the line .
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
We substitute y=0 in equation (found in (a)) of the line ;
Hence, coordinates of xintercept of the line (point T) are .
Coordinates of points P(5,6), and .
Now we can find distances SP and PT.










Hence;
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