The equation of one of the straight lines which passes through the point

Question:

The equation of one of the straight lines which passes through the point $(1,3)$ and makes an angles

$\tan ^{-1}(\sqrt{2})$ with the straight line, $y+1=3 \sqrt{2} x$ is

 

  1. $4 \sqrt{2} x+5 y-(15+4 \sqrt{2})=0$

  2. $5 \sqrt{2} x+4 y-(15+4 \sqrt{2})=0$

  3. $4 \sqrt{2} x+5 y-4 \sqrt{2}=0$

  4. $4 \sqrt{2} x-5 y-(5+4 \sqrt{2})=0$


Correct Option: 1

Solution:

$y=m x+c$

$3=m+c$

$\sqrt{2}=\left|\frac{m-3 \sqrt{2}}{1+3 \sqrt{2} m}\right|$

$=6 m+\sqrt{2}=m-3 \sqrt{2}$

$=\sin =-4 \sqrt{2} \rightarrow m=\frac{-4 \sqrt{2}}{5}$

$==6 m-\sqrt{2}=m-3 \sqrt{2}$

$=7 m-2 \sqrt{2} \rightarrow m=\frac{2 \sqrt{2}}{7}$

According to options take $\mathrm{m}=\frac{-4 \sqrt{2}}{5}$

So $\mathrm{y}=\frac{-4 \sqrt{2} \mathrm{x}}{5}+\frac{3+4 \sqrt{2}}{5}$

$4 \sqrt{2} x+5 y-(15+4 \sqrt{2})=0$

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now