(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)
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°