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Question 1 AQA AS Maths Pure Core 1, May 2011


Well-known member
Jan 26, 2012
1. The line \(AB\) has equation \(7x+3y=13\).

(a) Find the gradient of \(AB\). (2 marks)

(b) The point \(C\) has coordinates \((-1,3)\).

(i) Find an equation of the line which passes through the point \(C\) and which is parallel to \(AB\). (2 marks)

(ii) The point \( (1\frac{1}{2},-1)\) is the mid point of \(AC\). Find the coordinates of the point \(A\) (2 marks)

(c) The line \(AB\) intersects the line with equation \(3x+2y=12\) at the point \(B\). Find the coordinates of \(B\) (3 marks)


Well-known member
Jan 26, 2012
Re: Question 1 AQA AS Maths Pure Core 1, May 2011 Solution

(a) We rewrite the equation of the line in the form \(y=mx+c\), then \(m\) is the gradient:
\[3y=-7x+13 \\ y=-\;\frac{7}{3}x+\frac{13}{3}\],
so the gradient is \(-\frac{7}{3}\)

(b)(i) The line through \(C: (-1,3)\) has gradient \(-\frac{7}{3}\), and so is of the form: \( y=-\;\frac{7}{3}x+c\), and as it passes through \(C\) we have:\[3=-\;\frac{7}{3}\times (-1)+c\], so \(c=3-\frac{7}{3}=\frac{2}{3}\) and the required equation of the lone through \(C\) parallel to \(AB\) is:\[y=-\;\frac{7}{3}x+\frac{2}{3}\]. Or multipling through by \(3\):\[3y=-7x+2\]

(b)(ii) As \(M: (1\frac{1}{2},-1) \) is the mid-point of \(AC\) we have the coordinates of \(M\) are the average of the coordinates of \(A\) and \(C\), so if \(A: (a_1,a_2)\) we have:\[\begin{aligned} 3/2&=(a_1+(-1))/2 \\ (-1)&=(a_2+3)/2 \end{aligned}\]. Hence \(a_1=4\) and \(a_2=-5\), so \(A\) is the point \((4,-5)\)

(c) As the line \(AB\) intersects the line \(3x+2y=12\) at \(B\) we may substitute \(y\) from the equation for \(AB\) into this to get: \[3x+2\left(-\frac{7}{3}x+\frac{13}{3}\right)=12\]or \[\left( 3-\frac{14}{3}\right)x+\frac{26}{3}=12\]simplifying some more: \[\left( -\;\frac{5}{3}\right)x=\frac{10}{3}\] so \(x=-2\), and substituting back into either equation gives \(y=9\) so \(B\) is the point \((-2,9)\).

Note I have left out the checking of answers as you go along, it takes very little time to substitute values back into equations they are supposed to satisfy to confirm that they do indeed do so, etc.
Last edited:

The Chaz

Jan 26, 2012
Sorry for the necroposting...


"Point-slope form": $$y-y_1 = m(x-x_1)$$

I recommend this when they say "find **an** equation...", as there are infinitely many such equations (and some are easier to find than others!)

So $$y - 3 = \dfrac{-7}{3} (x - (-1))$$, or
$$ y -3= \dfrac{-7}{3} (x + 1)$$