# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2013 | January | Q#5

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**Question**

The line L_{1} has equation y = -2x+3.

The line L_{2} is perpendicular to L_{1} and passes through the point (5, 6).

**a. **Find an equation for L_{2} in the form ax + by + c = 0, where a, b and c are integers.

The line L_{2} crosses the x-axis at the point A and the y-axis at the point B.

**b. **Find the x-coordinate of A and the y-coordinate of B.

Given that O is the origin,

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

**Solution**

**a.
**

We are required to find equation of line L_{2}.

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 (5,6) on line L_{2} but not slope of the line L_{2}.

Next we need slope of the line L_{2 }to write its equation.

We are given that the line L_{2} is perpendicular to L_{1}.

If two lines are perpendicular (normal) to each other, then product of their slopes and is;

Therefore, if we have slope of line L_{1}, we can find slope of the line L_{2}.

We do not have slope of both lines L_{1 }and L_{2}. However, we are given equation of line L_{1} as;

Slope-Intercept form of the equation of the line;

Where is the slope of the line.

Therefore;

Hence;

With coordinates of a point on the line L_{2 }as (5,6) and its slope at hand, we can write equation of the line L_{2}.

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

**b.
**

We are given that the line L_{2} crosses the x-axis at the point A and the y-axis at the point B and we are required to find the x-coordinate of A and the y-coordinate of B.

It is evident that we are looking for coordinates of x-intercept (point A) and y-intercept (point B) of the line L_{2}.

First we find coordinates of point A (xintercept).

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 in (a) that equation of the line L_{2} is;

We substitute ;

Hence, x-coordinate of point A is “-7”.

Next we find coordinates of point B (y-intercept)

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 have found in (a) that equation of the line L_{2} is;

We substitute ;

Hence, y-coordinate of point B is “”.

**c. **

We are required to find the area of the triangle OAB.

We draw all these points and subsequent triangle.

Expression for the area of the triangle is;

From the diagram above, we can write;

It is evident from the diagram that OA is the horizontal distance of point A from origin O and that is 7 units. Similarly, OB is the vertical distance of point B from origin O and that is units.

Therefore;

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