Find the angles of a triangle whose sides are given by the lines 3x+y-1=0,
x – 3y + 7 = 0 and x + 2y – 8 = 0.
Answers
Answer:
The three angles are 90°, 45° and 45°.
Step-by-step explanation:
The three lines are
3x + y - 1 = 0 or, y = (- 3)x + 1 .....(1)
x - 3y + 7 = 0 or, y = (1/3)x + 7/3 .....(2)
x + 2y - 8 = 0 or, y = (- 1/2)x - 4 .....(3)
The gradient of the lines (1), (2) and (3) are (- 3), 1/3 and (- 1/2) respectively.
The angle between the lines (1) and (2) is
= tan⁻¹ | {(- 3) - (1/3)}/{1 + (- 3)(1/3)} |
= tan⁻¹ | (- 3 - 1/3)/(1 - 1) |
= tan⁻¹ | ( - 10/3)/0 |
= 90°
the angle between the lines (2) and (3) is
= tan⁻¹ | {1/3 - (- 1/2)}/{1 + (1/3)(- 1/2)} |
= tan⁻¹ | (1/3 + 1/2)/(1 - 1/6) |
= tan⁻¹ | (5/6)/(5/6) |
= tan⁻¹ (1)
= 45° and
the angle between the lines (3) and (1) is
= tan⁻¹ | {(- 1/2) - (- 3)}/{1 + (-1/2)(-3)} |
= tan⁻¹ | (- 1/2 + 3)/(1 + 3/2) |
= tan⁻¹ | (5/2)/(5/2) |
= tan⁻¹ (1)
= 45°
∴ three angles of the triangle are
90°, 45° and 45°