Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2011 | January | Q#9

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Question

The line L1 has equation 2y − 3x − k = 0, where k is a constant.

Given that the point A (1, 4) lies on L1, find

a.   the value of k,

b.   the gradient of L1.

The line L2 passes through A and is perpendicular to L1.

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

The line L2 crosses the x-axis at the point B.

d.   Find the coordinates of B.

e.   Find the exact length of AB.

Solution

a.
 

We are given that line L1 has equation;

We are given that the point A (1, 4) lies on L1.

If a point P(x,y) lies on a line, then its coordinates satisfy the equation of the line.

Therefore, we substitute coordinates of point A (1,4) in equation of L1.

b.
 

We are required to find the gradient of line L1.

Slope-Intercept form of the equation of the line;

Where  is the slope of the line.

We can rearrange the equation of the line in slope-intercept for to find the gradient.

From (a) we have found that k=5;

Hence, gradient of line L1 is;

c.
 

We are required to find equation of 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 are given that line L2 passes through the point A(1,4) and is perpendicular to line L1

If a line  is normal to the curve , then product of their slopes  and  at that point (where line  is normal to the curve) is;

Therefore;

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

From (a) we have found that;

Therefore;

Now we can write the equation of line L2.

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

Therefore;

d.
 

We are given that line L2 crosses x-axis at point B which means point B is xintercept of line L2

We are required to find the coordinates of x-intercept of line L2.

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).

Therefore;

Hence, coordinates of point B are (7,0).

 

e.
 

We are required to find the exact length of AB.

Expression to find distance between two given points  and is:

We are given coordinates of point A as (1,4) and we have found coordinates of point B in (d) as  (7,0).

Therefore;

Since ;

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