Past Papers’ Solutions  Assessment & Qualification Alliance (AQA)  AS & A level  Mathematics 6360  Pure Core 1 (6360MPC1)  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 (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope form of Equation of Line).
We already have coordinates of two points on the line A(1,2) and B(3,5).
TwoPoint form of the equation of the line is;
For line AB;
b.
i.
To find the midpoint of a line we must have the coordinates of the endpoints of the line.
Expressions for coordinates of midpoint of a line joining points and;
xcoordinate of midpoint of the line
ycoordinate of midpoint 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;
xcoordinate of midpoint of the line
ycoordinate of midpoint 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 (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope 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.
PointSlope 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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