Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2009 | January | Q#10

 

Question

The line  passes through the point A (2, 5) and has gradient .

a.  Find an equation of , giving your answer in the form y = mx + c.

The point B has coordinates (–2, 7).

b.   Show that B lies on .

c.   Find the length of AB, giving your answer in the form , where k is an integer.

The point C lies on  and has x-coordinate equal to p.

The length of AC is 5 units.

d.   Show that p satisfies

Solution

a.
 

We are required to find equation of .

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  passes through the point A (2, 5) and has gradient

Now we can write the equation of line .

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

Therefore;

b.
 

We are required to show that the point B (–2, 7) lies on line .

If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve  (or the line).

We have found in (a) that equation of line  is; 

We substitute x-coordinate of point B in the equation of the line  is;

Since coordinates of the point B satisfy the equation of the line , therefore the point lies on the  line.

c.
 

We are required to find the length AB.

Expression to find distance between two given points  and is:

We are given coordinates of both points A(2,5) and B(-2,7).

Since ;

d.

We are also given that length of AC is 5 units.

Expression to find distance between two given points  and is:

Therefore;

We are given coordinates of point A(2,5). But we are not given coordinates of point C which lies on   with x-coordinate p.

If we are given x-coordinate of point C, we can find y-coordinate by substituting x-coordinate of  point C in equation of the line  (on which point C lies).

From (a) we have equation of the line ;

Therefore, coordinates of point

Hence;

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