On a graph sheet, using a scale of 2cm to 2 units on both axes, (a) Draw the straight line joining points P(-5, 3) and Q(2, 3); (b) construct the locus L of...

The scale gives \(1\) coordinate unit \(=1\text{ cm}\). Therefore \(PQ=7\text{ cm}\), since \(P(-5,3)\) and \(Q(2,3)\) have the same \(y\)-coordinate.

Locus \(L\): Points equidistant from \(P\) and \(Q\) lie on the perpendicular bisector of \(PQ\). The midpoint of \(PQ\) is

\[ \left(\frac{-5+2}{2},\frac{3+3}{2}\right)=(-1.5,3). \]

Since \(PQ\) is horizontal, its perpendicular bisector is the vertical line \(x=-1.5\).

Construction of \(R\) and \(S\): Draw arcs of radius \(5\text{ cm}\), centred at \(P\) and at \(Q\). Their two intersections are \(R\) and \(S\). Each intersection is \(5\text{ cm}\) from both \(P\) and \(Q\), so \(PR=RQ=QS=SP=5\text{ cm}\), giving a rhombus.

Half of \(PQ\) is \(3.5\text{ cm}\). Using Pythagoras in the right triangle from the midpoint of \(PQ\) to \(R\):

\[ \text{vertical distance}=\sqrt{5^2-3.5^2} =\sqrt{25-12.25} =\sqrt{12.75} =\frac{\sqrt{51}}{2}\approx3.57. \]

Therefore, the coordinates are

\[ R=\left(-1.5,\;3+\frac{\sqrt{51}}{2}\right)\approx(-1.5,6.57), \]

\[ S=\left(-1.5,\;3-\frac{\sqrt{51}}{2}\right)\approx(-1.5,-0.57). \]

The diagonal \(PQ=7\text{ cm}\), and the other diagonal is

\[ RS=2\left(\frac{\sqrt{51}}{2}\right)=\sqrt{51}\text{ cm}. \]

Hence the area is

\[ \text{Area}=\frac{1}{2}\times PQ\times RS =\frac{1}{2}\times7\times\sqrt{51} \approx24.74\text{ cm}^2. \]

The supplied reference coordinates \((-1.6,6.5)\) and \((-1.6,0.6)\) are not consistent with the perpendicular bisector \(x=-1.5\), and the positive \(y\)-coordinate for \(S\) is incorrect. A construction read from a graph may reasonably give approximately \(R(-1.5,6.6)\) and \(S(-1.5,-0.6)\).