Question

Find the distance of the point (2, 5) from the line 3 x + y + 4 = 0 measured parallel to a line having slope $3/4.$

Answer 1

$\text{Here}(x_1,y_1)=A(2,5)and$

$\text{Slope of the line}=tan\alpha =\frac{3}{4}$

$\therefore sin\alpha =\frac{3}{\sqrt{3^2+4^2}}=\frac{3}{5}\text{and}cos\alpha =\frac{4}{\sqrt{3^2+4^2}}=\frac{4}{5}$

$\therefore \text{Equation of line is passing through}A(2,5)\text{and having slope}\frac{3}{4}is$

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

$\frac{x-2}{cos\alpha }=\frac{y-5}{sin\alpha }=r$

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

$orx=\frac{4r}{5}+2andy=\frac{3r}{5}+5$

$thenP(\frac{4r}{5}+2)+(\frac{3r}{5}+5)\text{lie on}$

$3x+y+4=0$

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

$\frac{12r}{5}+6+\frac{3x}{5}+5+4=0$

$\frac{15r}{5}+15=0$

$\frac{15}{5}r=-15$

$r=-\frac{15\times 5}{15}$

$=5units$

$\text{Hence the distance of the point (2, 5)}$

$\text{from the line}3x+y+4=0is5$