Let []1, []2 ([]1 2) be the values of [] fo the points ([], -3), (2, 0) and (1, []) to be …

Question

Let $$\alpha$$1, $$\alpha$$2 ($$\alpha$$1 2) be the values of $$\alpha$$ fo the points ($$\alpha$$, -3), (2, 0) and (1, $$\alpha$$) to be collinear. Then the equation of the line, passing through ($$\alpha$$1, $$\alpha$$2) and making an angle of $${\pi  \over 3}$$ with the positive direction of the x-axis, is :

Accepted Answer

Points A($$\alpha$$, -3), B(2, 0) and C(1, $$\alpha$$) are collinear.

$$\therefore$$ Slope of AB = Slope of BC

$$ \Rightarrow {{0 + 3} \over {2 - \alpha }} = {{\alpha  - 0} \over {1 - 2}}$$

$$ \Rightarrow  - 3 = \alpha (2 - \alpha )$$

$$ \Rightarrow  - 3 = 2\alpha  - {\alpha ^2}$$

$$ \Rightarrow {\alpha ^2} - 2\alpha  - 3 = 0$$

$$ \Rightarrow {\alpha ^2} - 3\alpha  + \alpha  - 3 = 0$$

$$ \Rightarrow \alpha (\alpha  - 3) + 1(\alpha  - 3) = 0$$

$$ \Rightarrow (\alpha  + 1)(\alpha  - 3) = 0$$

$$ \Rightarrow \alpha  =  - 1,\,3$$

Given, $${\alpha _1} < {\alpha _2}$$

$$\therefore$$ $${\alpha _1} = -1$$ and $${\alpha _2} = 3$$

$$\therefore$$ $$\left( {{\alpha _1},\,{\alpha _2}} \right) = ( - 1,\,3)$$

Now, equation of the line passing through (-1, 3) and making angle $${\pi  \over 3}$$ with positive x-axis is

$(y - {y_1}) = m(x - {x_1})$

$$ \Rightarrow y - 3 = \left( {\tan {\pi  \over 3}} \right)(x + 1)$$

$$ \Rightarrow y - 3 = \sqrt 3 (x + 1)$$

$$ \Rightarrow \sqrt 3 x - y + \sqrt 3  + 3 = 0$$

Solutions · 1