(a) Find the equation of the line passing through the points (2, 5) and (-4, -7). (b) Three ships P, Q and R are at sea. The bearing of Q from P is 030° and...

(a) Equation of the line

The gradient of the line through \((2,5)\) and \((-4,-7)\) is

\[ m=\frac{-7-5}{-4-2}=\frac{-12}{-6}=2. \]

Using the point \((2,5)\),

\[ y-5=2(x-2). \]

\[ y-5=2x-4 \]

\[ \boxed{y=2x+1}. \]

(b)(i) Illustration

As shown in the diagram above, \(Q\) is on a bearing of \(030^\circ\) from \(P\), while \(R\) is on a bearing of \(300^\circ\) from \(P\). Therefore, the angle between \(PQ\) and \(PR\) is

\[ \angle QPR=30^\circ+60^\circ=90^\circ. \]

Thus, \(\triangle PQR\) is right-angled at \(P\), with \(PQ=5\text{ km}\) and \(PR=8\text{ km}\).

(b)(ii)(1) Distance between \(Q\) and \(R\)

By Pythagoras' theorem,

\[ QR^2=PQ^2+PR^2 \]

\[ QR^2=5^2+8^2=25+64=89. \]

\[ QR=\sqrt{89}=9.433\ldots \]

\[ \boxed{QR=9.43\text{ km}} \]

correct to three significant figures.

(b)(ii)(2) Bearing of \(R\) from \(Q\)

First, find the angle at \(Q\):

\[ \tan \angle PQR=\frac{PR}{PQ}=\frac{8}{5}. \]

\[ \angle PQR=\tan^{-1}\left(\frac{8}{5}\right)=58.0^\circ. \]

The bearing of \(P\) from \(Q\) is the reverse bearing of \(030^\circ\):

\[ 030^\circ+180^\circ=210^\circ. \]

From the diagram, \(R\) lies clockwise from the direction \(QP\). Hence,

\[ \text{Bearing of }R\text{ from }Q=210^\circ+58^\circ=268^\circ. \]

\[ \boxed{268^\circ} \]