Question 1 AQA AS Maths Pure Core 1, May 2011

In summary: We can find the midpoint of two coordinates by finding the average of each coordinate.If the two points are \((x_1,y_1)\) and \((x_2,y_2)\), the midpoint is $$\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$$In this case, we have two points: \(A\) and \(C\).The coordinates of \(C\) are \((-1,3)\).The midpoint, or average, of the coordinates of \(C\) and \(M\) is $$\left(\dfrac{-1+1\frac{1}{2}}{
  • #1
CaptainBlack
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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)
 
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  • #2
Re: Question 1 AQA AS Maths Pure Core 1, May 2011 Solution

Answer:
(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.
 
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  • #3
Sorry for the necroposting...

1.b.i

"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
$$ y -3= \dfrac{-7}{3} (x + 1)$$
 

FAQ: Question 1 AQA AS Maths Pure Core 1, May 2011

What topics were covered in Question 1 of the May 2011 AQA AS Maths Pure Core 1 exam?

The topics covered in Question 1 were algebraic manipulation and solving equations, including factorising, expanding brackets, and solving linear and quadratic equations.

What was the difficulty level of Question 1 in the May 2011 AQA AS Maths Pure Core 1 exam?

The difficulty level of Question 1 was considered to be moderate. It required a good understanding of basic algebraic concepts and the ability to apply them to solve equations.

Were any calculators allowed for Question 1 in the May 2011 AQA AS Maths Pure Core 1 exam?

No, calculators were not allowed for any part of the AS Maths Pure Core 1 exam, including Question 1.

How many marks was Question 1 worth in the May 2011 AQA AS Maths Pure Core 1 exam?

Question 1 was worth a total of 20 marks, making it the highest weighted question on the exam paper.

Is there a specific method or strategy that should be used to answer Question 1 in the May 2011 AQA AS Maths Pure Core 1 exam?

There is no specific method or strategy that must be used for Question 1. However, it is important to read the question carefully, show all working, and check answers for accuracy.

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