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 L1 has equation y = -2x+3.

The line L2 is perpendicular to L1 and passes through the point (5, 6).

a.   Find an equation for L2 in the form ax + by + c = 0, where a, b and c are integers.

The line L2 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 L2.

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 L2 but not  slope of the line L2.

Next we need slope of the line Lto write its equation.

We are given that the line L2 is perpendicular to L1.

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

Therefore, if we have slope of line L1, we can find slope of the line L2.

We do not have slope of both lines Land L2. However, we are given equation of line L1 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 Las (5,6) and its slope   at hand, we can write equation of the line L2.

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

b.
 

We are given that the line L2 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 L2

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 L2 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 L2 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.

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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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