# 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.        