The equation of the line through the point $(-1,3)$ in symmetrical form, when the angle made by the line…

The equation of the line through the point $(-1,3)$ in symmetrical form, when the angle made by the line with the positive direction of $X$-axis is $120^{\circ}$, is given by
  1. $\frac{(x+1)}{-1 / 2}=\frac{(y-3)}{\sqrt{3} / 2}=r$
  2. $\frac{(x+1)}{1 / 2}=\frac{(y+3)}{\sqrt{3} / 2}=r$
  3. $\frac{(x+1)}{-1 / 2}=\frac{(y+3)}{\sqrt{3} / 2}=r$
  4. $\frac{(x+1)}{1 / 2}=\frac{(y-3)}{\sqrt{3} / 2}=r$

Solution

The equation of line through the point $\left(x_1, y_1\right)$ in symmetrical form, if it's inclination with positive direction of $X$-axis is $\theta$ is $ \frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r $ So, the equation of the line in symmetric form where $\left(x_1, y_1\right)=(-1,3)$ and $\theta=120^{\circ}$ is $ \frac{x+1}{\cos 120^{\circ}}=\frac{y-3}{\sin 120^{\circ}}=r \Rightarrow \frac{x+1}{-1 / 2}=\frac{y-3}{\sqrt{3} / 2}=r $

Asked in: AP EAMCET 2020 (22 Sep Shift 1)

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