Question

Find the distance of the point (2, 5) from the line 3x + y + 4 = 0 measured parallel to the line 3x - 4y + 8 = 0.

Answer 1

$\text{Here}(x_1,y_1)=A(2,5)\text{Slope of required line}$

$\text{It is given that required line is parallel to}$

$3x-4y+8=0$

$\Rightarrow 4y=3x+8$

$\Rightarrow y=\frac{3}{4}x+2$

$\therefore tan\theta =\frac{3}{4}$

$\Rightarrow sin\theta =\frac{3}{5},cos\theta =\frac{4}{5}$

$\therefore \text{Equation of required line}$

$\frac{x-x_1}{cos\theta }=\frac{y-y_1}{sin\theta }=r$

$\frac{x-2}{\frac{4}{5}}=\frac{y-5}{\frac{3}{5}}=r$

$or$

$p(\frac{4}{5}r+2,\frac{3r}{5}+5)$

$\text{and P lies in}3x+y+4=0$

$\therefore 3(\frac{4}{3}r+2)+(\frac{3r}{5}+5)+4=0$

$\Rightarrow 12r+30+3r+25+20=0$

$\Rightarrow 15r+75=0$

$\Rightarrow r=5$