(a) Find the equation of the line passing through the points (2, 5) and (-4,
(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 the bearing of P and R is 300°. If |PQ| = 5 km and |PR| = 8 km,
(i) Illustrate the information in a diagram.
(ii) Calculate, correct to three significant figures, the:
(1) distance between Q and R
(2) bearing of R from Q.
Explanation
(a) Using the two- point form,
\(\frac{y - y_{1}}{y_{2} - y_{1}} = \frac{x - x_{1}}{x_{2} - x_{1}}\)
\(\frac{y - 5}{-7 - 5} = \frac{x - 2}{-4 - 2}\)
\(\frac{y - 5}{- 12} = \frac{x - 2}{- 6}\)
\(y - 5 = 2(x - 2)\)
\(y - 5 = 2x - 4 \implies y - 2x = -4 + 5 = 1\)
\(Equation : y = 2x + 1\)
(b) (i).jpg)
In \(\Delta PQR, <QPR = 60° + 30° = 90°\). PQR is a right- angled triangle.
(ii) (1) In \(\Delta PQR, |QR|^{2} = 5^{2} + 8^{2}\)
\(25 + 64 = 89\)
\(|QR| = \sqrt{89} = 9.434 km \approxeq 9.43 km\)
(2) In the diagram above, the bearing of R from Q is the obtuse angle NQR.
But \(\tan \alpha = \frac{8}{5} = 1.6\)
\(\alpha = \tan^{-1} (1.6) = 58°\)
Hence, angle NQR = 360° - (a + 60° + 90°)
= 360° - (58° + 60° + 90°)
= 360° - 208°
= 152°