\includegraphics{figure_2}
Figure 2 is a sketch showing the line \(l_1\) with equation \(y = 2x - 1\) and the point \(A\) with coordinates \((-2, 3)\).
The line \(l_2\) passes through \(A\) and is perpendicular to \(l_1\)
- Find the equation of \(l_2\) writing your answer in the form \(y = mx + c\), where \(m\) and \(c\) are constants to be found. [3]
The point \(B\) and the point \(C\) lie on \(l_1\) such that \(ABC\) is an isosceles triangle with \(AB = AC = 2\sqrt{13}\)
- Show that the \(x\) coordinates of points \(B\) and \(C\) satisfy the equation
$$5x^2 - 12x - 32 = 0$$ [4]
Given that \(B\) lies in the 3rd quadrant
- find, using algebra and showing your working, the coordinates of \(B\). [4]