# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2012 | January | Q#6

**Question**

The line has equation 2x − 3y +12 = 0.

**a. **Find the gradient of .

The line crosses the x-axis at the point A and the y-axis at the point B, as shown in Figure.

The line is perpendicular to and passes through B.

**b. **Find an equation of .

The line crosses the x-axis at the point C.

**c. **Find the area of triangle ABC.

**Solution**

**a. **

We are given equation of line ;

We are required 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;

**b. **

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 neither coordinates of a point on line nor slope of the line .

First we find coordinates of a point on line .

We are given that line passes through B. We are also given that the line crosses the y-axis at the point B. Therefore, point B is 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 are given equation of line ;

We substitute ;

Hence, coordinates of point B are (0,4).

Now we have coordinates of a point on 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 .

From (a), we have found that , therefore;

With coordinates of a point on the line as B(0,4) and its slope

at hand, we can write equation of the line .

Point-Slope form of the equation of the line is;

**c.
**

We are required to find the area of .

We need coordinates of all three vertexes of triangle to sketch it in order to calculate its area.

We have found in (b) that coordinates of point B are (0,4).

We now find coordinates of point A.

We are given that the line crosses the x-axis at the point A. Therefore, point A 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 are given equation of the line ;

We substitute ;

Therefore, coordinates of point A are (-6,0).

Next, we find coordinates of point C.

We are given that the line crosses the x-axis at the point C. Therefore, point C 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 have found equation of the line in (b);

We substitute ;

Therefore, coordinates of point C are .

We sketch triangle with the help of these coordinates of three points.

Expression for the area of the triangle is;

For ;

We need to find AC and BD.

First we find AC.

Expression for the distance between two given points and is:

We have coordinates of and . Therefore;

Next we need to find BD. It is evident from the diagram that it is only the distance of B from origin D(0,0) along y-axis which is 4 since B(4,0).

Therefore;

Hence;

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