# 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 L_{1} has equation 2y − 3x − k = 0, where k is a constant.

Given that the point A (1, 4) lies on L_{1}, find

**a. **the value of k,

**b. **the gradient of L_{1}.

The line L_{2} passes through A and is perpendicular to L_{1}.

**c. **Find an equation of L_{2} giving your answer 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 B.

**d. **Find the coordinates of B.

**e. **Find the exact length of AB.

**Solution**

**a.
**

We are given that line L_{1} has equation;

We are given that the point A (1, 4) lies on L_{1}.

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 L_{1}.

**b.
**

We are required to find the gradient of line L_{1}.

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 L_{1} is;

**c.
**

We are required to find equation of 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 are given that line L_{2} passes through the point A(1,4) and is perpendicular to line L_{1}.

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 L_{1} we can find slope of line L_{2}.

From (a) we have found that;

Therefore;

Now we can write the equation of line L_{2}.

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

Therefore;

**d.
**

We are given that line L_{2} crosses x-axis at point B which means point B is xintercept of line L_{2}.

We are required to find the coordinates of x-intercept of line L_{2}.

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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