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Question

(A) The equation of a line passing through the point $(x_1, y_1)$ in symmetrical form with an inclination $	heta$ to the positive $X$-axis is given b

Vedclass · Accepted Answer

$\frac{x+1}{-1/2} = \frac{y-3}{\sqrt{3}/2} = r$

Vedclass · Answer

(A) The equation of a line passing through the point $(x_1, y_1)$ in symmetrical form with an inclination $	heta$ to the positive $X$-axis is given by: $\frac{x-x_1}{\cos 	heta} = \frac{y-y_1}{\sin 	heta} = r$ Given $(x_1, y_1) = (-1, 3)$ and $	heta = 120^{\circ}$,we have: $\cos 120^{\circ} = -1/2$ and $\sin 120^{\circ} = \sqrt{3}/2$ Substituting these values into the formula: $\frac{x - (-1)}{-1/2} = \frac{y - 3}{\sqrt{3}/2} = r$ $\frac{x+1}{-1/2} = \frac{y-3}{\sqrt{3}/2} = r$

The equation of the line passing through the point $(-1, 3)$ in symmetrical form,when the angle made by the line with the positive direction of the $X$-axis is $120^{\circ}$,is given by:

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