Question

236. The angle between the lines x - 2y = 2
and y-2x = 5 is
(A) tan' (1/4)
(B) tan (35)
(C) tan(5/4)
(D) tan' (2/3)​

Answer 1

EXPLANATION.

Angles between the lines x - 2y = 2.

y - 2x = 5.

→ Slope of line y = mx + c.

→ Slope of line x - 2y = 2

→ 2y = 2 - x

→ y = 2/2 - x/2.

→ Slope = -1/2.

→ Slope of line y - 2x = 5.

→ y = 2x + 5

→ Slope = 2.

$\sf \: \implies \: \tan( \theta) = | \dfrac{ m_{1} \: - \: m_{2} }{1 \: + \: m_{1} m_{2} } | \\ \\ \sf \: \implies \: \tan( \theta) = \: | \dfrac{ \frac{ - 1}{2} - 2 }{1 + \dfrac{ - 1}{2} \times 2 } | \\ \\ \sf \: \implies \: \tan( \theta) = | \dfrac{ \dfrac{ - 1 - 4}{2} }{1 - 1} | \\ \\ \sf \: \implies \: \tan( \theta) = 0$

Answer 2

Angles between the lines x - 2y = 2.

y - 2x = 5.

→ Slope of line y = mx + c.→ Slope of line x - 2y

= 2→ 2y

= 2 - x→ y

= 2/2 - x/2

$\leadsto \: \ \mathfrak{tan ( \theta) = | \frac{ m_{1} - m_{2} }{1 + m_{1} m_{2} } | } \\ \\ \\ \leadsto \mathfrak{tan ( \theta) = | \frac{ \frac{ - 1}{2} - 2 }{1 + \frac{ - 1}{2} \times 2 } | } \\ \\ \\ \leadsto \mathfrak{ tan( \theta) = \frac{ \frac{ - 1 - 4}{2} }{ |1 - 1| } } \\ \\ \\ \leadsto \mathfrak{ tan( \theta) = 0}$