# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2014 | June | Q#9

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**Question**

The line , shown in figure has equation 2x+3y = 26.

The line passes through the origin O and is perpendicular to .

**a. **Find an equation for the line .

The line intersects the line at the point C.

Line crosses the y-axis at the point B as shown in Figure.

**b. **Find the area of triangle OBC.

Give your answer in the form , where a and b are integers to be determined.

**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 coordinates of a point O(0,0) on line but not slope of the line .

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 O(0,0) 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 OBC.

Expression for the area of the triangle with base and height is;

Consider the figure below.

It is evident that OB can be considered as base of triangle OBC whereas AC is height of the triangle.

We need to find distances OB and AC.

Expression for the distance between two given points and is:

We do not have coordinates of these points so let us find them first.

We have coordinates of O(0,0).

We need to find coordinates of point B, next.

Point B is the y-intercept of the line .

The point at which curve (or line) intercepts y-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).

We substitute x=0 in equation of the line ;

Hence, coordinates of y-intercept of the line (point B) are .

Next we need coordinates of point C which is the point of intersection of 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).

Equation of the line is;

Equation of the line (as found in (a)) is;

We need to solve these simultaneous equations.

Substitute expression of y from second equation in first;

.

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 equation of the line ;

Substitute;

Hence, coordinates of point of intersection of lines C are (4,6).

Now we can find distances OB and AC.

First we can see for OB that it is only vertical (along y-axis) distance of point B from O(0,0), therefore;

Similarly, AC is only horizontal (along x-axis) distance of point C(4,6) from point A, therefore;

Hence;

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