# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2017 | June | Q#8

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 x-axis 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 (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 have x-coordinates of a point P(5,y) on line  but not slope of the line .

First we find y-coordinate of the point P on the line . Since this point also lies on the line , we   can utilize equation of the line  to find y-coordinate 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 .

Slope-Intercept form of the equation of the line;

Where  is the slope of the line.

Therefore, we can rearrange the given equation of line  in slope-intercept 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

Point-Slope 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 x-intercept of the line .

The point  at which curve (or line) intercepts x-axis, 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 x-intercept of the line  (point S) are .

Next we need coordinates of point T which is the x-intercept of the line .

The point  at which curve (or line) intercepts x-axis, 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 x-intercept of the line  (point T) are .

Coordinates of points P(5,6),  and

Now we can find distances SP and PT.

Hence;