Question

find inclination made by the line joining points (2a,2aroot3) and (3a,3aroot3) in direction of X-axis?​

Answer 1

$\begin{gathered}\begin{gathered}\bf \: Given \: coordinates \: are \: \begin{cases} &\sf{A(2a, 2a \sqrt{3} )} \\ &\sf{B(3a, 3a \sqrt{3} )} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \:To\:find - \begin{cases} &\sf{angle \: makes \: with \: direction \: of \: x - axis}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline{\bold{Solution-}}$

• Let suppose that line AB makes an angle θ with positive direction of x - axis.

We know,

• If line passes through two points A (a, b) and B (c, d) and makes an angle θ with positivedirectionofx - axis, then slope (m) of line is

$\sf \: m \: = \: \dfrac{d - b}{c - a} = tan \theta \:$

Here,

$\: \: \: \: \: \: \bull \: \sf \: a \: = \: 2a$

$\: \: \: \: \: \: \bull \: \sf \: b \: = \: 2a \sqrt{3}$

$\: \: \: \: \: \: \bull \: \sf \: c \: = \: 3a$

$\: \: \: \: \: \: \bull \: \sf \: d \: = \: 3a \sqrt{3}$

So, on substituting the values, we get

$\bf \: tan \theta \: = \rm \: \dfrac{3a \sqrt{3} - 2a \sqrt{3} }{3a - 2a}$

$\bf \: tan \theta \: = \rm \: \dfrac{a \sqrt{3} }{a}$

$\bf \: tan \theta \: = \: \rm \: \sqrt{3}$

$\bf\implies \: \theta \: = \: 60 \degree$

Additional Information

1. Two lines having slope m and M are parallel iff m = M.

2. Two lines having slope m and M are perpendicular iff mM = - 1

3. If line is parallel to x - axis, its slope is 0.

4. If a line is parallel to y - axis, its slope is not defined.

5. If 3 points A, B, C are collinear, then slope of AB = slope of BC.