Find the equations of the straight lines passing through (2, -1) … - Vidyadip

Question

Find the equations of the straight lines passing through $(2, -1)$ and making an angle of $45°$ with the line $6x + 5y - 8 = 0$.

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Answer

We know that the equations of two lines passing through a point $(x_1,y_1)$ and making an angle $\alpha$ with the given line y = mx + c are $\text{y}-\text{y}_1=\frac{\text{m}\pm\tan\alpha}{1\mp\text{m}\tan\alpha}(\text{x}-\text{x}_1)$ Here, Equation of the given line is, $6\text{x}+5\text{y}-8=0$
$\Rightarrow5\text{y}=-6\text{x}+8$
$\Rightarrow\text{y}=-\frac{6}{5}\text{x}+\frac{8}{5}$ Comparing this equation with y = mx + c we get, $\text{m}=-\frac{6}{5}$
$\text{x}_1=2,\text{ y}_1=-1,\alpha=45^\circ,\text{ m}=-\frac{6}{5}$ So, the equation of the required lines are $\text{y}+1=\frac{-\frac{6}{5}+\tan45^\circ}{1+\frac{6}{5}\tan45^\circ}(\text{x}-2)$ and $\text{y}+1=\frac{-\frac{6}{5}-\tan45^\circ}{1-\frac{6}{5}\tan45^\circ}(\text{x}-2)$
$\Rightarrow\text{y+1=}\frac{-\frac{6}{5}+1}{1+\frac{6}{5}}(\text{x}-2)$ and $\text{y+1}=\frac{-\frac{6}{5}-1}{1-\frac{6}{5}}(\text{x}-2)$
$\Rightarrow\text{y+1}=\frac{-1}{11}(\text{x}-2)$ and $\text{y+1}=\frac{-11}{-1}(\text{x}-2)$
$\Rightarrow\text{x}+11\text{y}+9=0$ and $11\text{x}-\text{y}-23=0$

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