Question

The slope of a line is double of the slope of another line.If tangent of the angle b/w them is 1/3 , find the slopes of the line?

Answer 1

|(M2-m1)/(1+m1M2)| = tan x , x being d angle bw the lines and M2 and m1 being the slopes

Answer 2

$\huge \fbox{ \mathtt {SOLUTION :)</p><p>}}$

Let , the slope a another of line be m

So , slope of a line = 2m

Given , The tangent of the angle between two given line is 1/3

This implies,

$\mathtt{\fbox{Tan( \theta) = \frac{1}{3} }}$

We know that , the acute angle between two lines with slope m1 and m2 is given by

$\large\mathtt{\fbox{Tan( \theta) = | \frac{ m_{2} - m_{1} }{1 + m_{1} m_{2} } | }} - - (i)$

Let m1 = 2m , m2 = m and tan(theta) = 1/3

Now , putting these values in (i) , we get

{ if (m2-m1/1+m1m2) is positive } , then ,

$\sf \hookrightarrow \frac{1}{3} = | \frac{m - 2m}{1 + 2m(m)} | \\ \\\sf \hookrightarrow \frac{1}{3} = | \frac{ - m}{1 + 2 {m}^{2} } | \\ \\ \sf \hookrightarrow \frac{1}{3} = \frac{m}{1 + 2 {m}^{2} } \\ \\\sf \hookrightarrow 1 + 2 {m}^{2} = 3m \\ \\ \sf \hookrightarrow 2 {m}^{2} - 3m + 1 = 0 \\ \\ \sf \hookrightarrow 2 {m}^{2} - 2m - m + 1 = 0 \\ \\\sf \hookrightarrow 2m(m - 1) - 1(m - 1) = 0 \\ \\\sf \hookrightarrow (2m - 1)(m - 1) \\ \\ \sf \hookrightarrow m = \frac{1}{2} \: \: or \: \: m = 1$

Hence , the slope are 1/2 and 1 or 1 and 2 { if (m2-m1/1+m1m2) is positive

{ if (m2-m1/1+m1m2) is negative } ,

Then ,

$\sf \hookrightarrow m = - \frac{1}{2} \: \: or \: \: m = - 1$

You may get this answer by the same method as if (m2-m1/1+m1m2) is positive

Hence , the slope of the lines are -1/2 and -1 or -1 and -2 { if (m2-m1/1+m1m2) is negative }