Question

The triangle STV has its vertices at s ( 2,5).The angle TVS is a right angle and the equation of VS is 2x+3y -1=0

Find the distance of S from the line VT

if ST has a gradient of 1/3,find the angle between the lines ST and TV

Answer 1

$2\cdot 2+3\cdot 5-1=18\neq0 \Rightarrow (2,5)\not\in VS :\quad 2x+3y-1=0 \Rightarrow T=T(2,5)\\ ST:\quad y=\frac{1}{3}x+b, \quad T\in ST: \quad y=\frac{1}{3}x+b \Rightarrow b=\frac{13}{3}\\ S=\begin{cases} 2x+3y-1=0\\y=\frac{1}{3}x+\frac{13}{3} \end{cases} \Rightarrow S=S(-4,3).\\ TV: \quad y-5=\frac{-1}{\frac{-2}{3}}(x-2)=\frac{3}{2}(x-2).\\ cos(\alpha)=\frac{\frac{1}{3}\cdot\frac{3}{2}+(-1)\cdot(-1)}{\sqrt{\frac{1}{9}+1}\cdot \sqrt{\frac{9}{4}+1}}=\frac{9}{\sqrt{130}} \Rightarrow\alpha=arccos(\frac{9}{\sqrt{130}}).\\ \rho(S,TV)=\frac{|1.5\cdot(-4)-3+2|}{\sqrt{\frac{9}{4}+1}}=\frac{14}{\sqrt{13}};$