Question

Find the distance of the point (3, 5) from the line 2 x + 3 y = 14 measured parallel to a line having slope $1/2.$

Answer 1

$\text{Here,}(x_1,y_1)=A(3,5),tan\theta =\frac{1}{2}$

$\Rightarrow sin\theta =\frac{1}{\sqrt{1^2+2^2}}andcos\theta =\frac{2}{\sqrt{1^2+2^2}}$

$\Rightarrow sin\theta =\frac{1}{\sqrt{5}}andcos\theta =\frac{2}{\sqrt{5}}$

$\text{So, the equation of the line passing}$

$\text{through (3, 5) and having slope}\frac{1}{2}is$

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

$\Rightarrow \frac{x-3}{\frac{2}{\sqrt{5}}}=\frac{y-5}{\frac{1}{\sqrt{5}}}$

$\Rightarrow x-2y+7=0$

$Letx-2y+7=0\text{intersect the line}2x+3y=14\text{at point P.}$

$\text{Let AP = r}$

$\text{Then, the coordinates of P are given by}$

$\frac{x-3}{\frac{2}{\sqrt{5}}}=\frac{y-5}{\frac{1}{\sqrt{5}}}=r$

$\Rightarrow x=3+\frac{2r}{\sqrt{5}}andy=5+\frac{r}{\sqrt{5}}$

$\text{Thus, the coordinates of P are}$

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

$\text{Clearly, P lies on the line}2x+3y=14$

$\therefore 2(3+\frac{2r}{\sqrt{5}})+3(5+\frac{r}{\sqrt{5}})=14$

$\Rightarrow 6+\frac{4r}{\sqrt{5}}+15+\frac{3r}{\sqrt{5}}=14$

$\Rightarrow \frac{7r}{\sqrt{5}}=-7$

$\Rightarrow r=-\sqrt{5}$

$\text{Hence, the distance of the point (3, 5)}$

$\text{from the line}2x+3y=14is\sqrt{5}$

$\text{So, the coordinates of}{P}_{1}and{P}_{2}are(1+rcos\theta ,5+rsin\theta )and(1-rcos\theta ,5-rsin\theta ),\text{respectively.}$

$\text{Clearly,}{P}_{1}and{P}_2\text{lie on}5x-y-4=0$

$and3x+4y-4=0,\text{respectively}$

$\therefore 5(1+rcos\theta )-5-rsin\theta -4=0and$

$3(1-rcos\theta )+4(5-rsin\theta )-4=0$

$\Rightarrow r=\frac{4}{5cos\theta -sin\theta }and$

$r=\frac{19}{3cos\theta +4sin\theta }$

$\Rightarrow \frac{4}{5cos\theta -sin\theta }=\frac{19}{3cos\theta +4sin\theta }$

$\Rightarrow 95cos\theta -19sin\theta =12cos\theta +16sin\theta$

$\Rightarrow 83cos\theta =35sin\theta$

$\Rightarrow tan\theta =\frac{83}{35}$

$\text{Thus, the equation of the required line is}$

$\frac{y-5}{x-1}=tan\theta$

$\Rightarrow \frac{y-5}{x-1}=\frac{83}{35}$

$\Rightarrow 83x-35y+92=0$