(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)
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 \((-2,9)\).
