Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2014 | June | Q#1

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Question

The point A has coordinates (-1,2) and the point B has coordinates (3,-5).

a.    

                    i.       Find the gradient of AB.

                  ii.       Hence find an equation of the line AB, giving your answer in the form  , where p, q  and r are integers.

b.   Midpoint of AB is M.

                    i.       Find the coordinates of M.

                  ii.       Find an equation of the line which passes through M and which is perpendicular to AB. 

c.   The point C has coordinates (k,2k+3). Given that the distance from A to C is  find the two  possible values of the constant k.

Solution

a.
 

                            i.
 

Expression for slope (gradient) of a line joining points  and ;

                          ii.
 

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 already have coordinates of two points on the line A(-1,2) and B(3,-5).

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

For line AB;

b.
 

                            i.
 

To find the mid-point of a line we must have the coordinates of the end-points of the line.

Expressions for coordinates of mid-point of a line joining points  and;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

We can also find slope of line AB from given coordinates of point s A(-1,2) and B(3,-5).

Hence;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Hence coordinates of point M are .

 

                         ii.
 

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 already have coordinates of a point on the line

We need slope of this line to write its equation.

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;

We have found slope of line AB in (a:i) as;

Therefore;

Now we can write equation of the normal to the curve.

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

c.
 

Expression to find distance between two given points  and is:

For distance AC;

We are given coordinates of the point C (k,2k+3) and point A(-1,2).

We are also given that distance from A to C is .

Therefore;

We have the algebraic formula;

Now we have two options.

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