Question

Find the angles of the triangle whose sides are

x + y - 4 = 0, 2 x + y - 6 = 0

and

5 x + 3 y - 15 = 0

.

Accepted Answer

Given lines are $x + y - 4 = 0$ ........(i)

$2 x + y - 6 = 0$ ..........(ii)

$5 x + 3 y - 15 = 0$ ..........(iii)

We know that, angle between two lines with slopes $m_{1} & m_{2}$ is $\tan θ = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}|$ .

Slope of line $x + y - 4 = 0$ is $m_{3} = - 1$

Slope of line $2 x + y - 6 = 0$ is $m_{1} = \frac{- 2}{1} = - 2$

Slope of line $5 x + 3 y + 5 = 0$ is $m_{2} = \frac{- 5}{3}$

Angle between lines (i) and (ii) is $\tan θ_{1} = |\frac{- 2 - (- 1)}{1 + (- 2) (- 1)}|$ $= |\frac{- 2 + 1}{1 + 2}|$ $= |\frac{- 1}{3}|$

$\Rightarrow \tan θ_{1} = \frac{1}{3}$

Therefore, angle between lines (i) and (ii) is $θ_{1} = \tan^{- 1} (\frac{1}{3})$ .

Angle between lines (ii) and (iii) is $\tan θ_{2} = |\frac{- 2 + \frac{5}{3}}{1 + - 2 \times - \frac{5}{3}}|$ $= |\frac{- \frac{6 + 5}{3}}{\frac{3 + 10}{3}}|$ $= |\frac{- 1}{13}|$

$\Rightarrow \tan θ_{2} = \frac{1}{13}$

Therefore, angle between lines (ii) and (iii) is $θ_{2} = \tan^{- 1} (\frac{1}{13})$ .

Angle between lines (i) and (iii) is $\tan θ_{3} = |\frac{\frac{- 5}{3} + 1}{1 + (- \frac{5}{3}) (- 1)}|$ $= |\frac{\frac{- 2}{3}}{\frac{3 + 5}{3}}|$ $= \frac{1}{4}$

Therefore, angle between lines (i) and (iii) is $θ_{3} = \tan^{- 1} (\frac{1}{4})$ .

Andhra Pradesh Board Solutions for Chapter: The Straight Line, Exercise 4: Exercise 3(d)

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Chapter: The Straight Line, Exercise 4: Exercise 3(d)

Questions from Andhra Pradesh Board Solutions for Chapter: The Straight Line, Exercise 4: Exercise 3(d) with Hints & Solutions

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