Answer to Question #129148 in Analytic Geometry for Sym

Question #129148

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


1
Expert's answer
2020-08-17T19:17:12-0400

22+351=180(2,5)∉VS:2x+3y1=0T=T(2,5)ST:y=13x+b,TST:y=13x+bb=133S={2x+3y1=0y=13x+133S=S(4,3).TV:y5=123(x2)=32(x2).cos(α)=1332+(1)(1)19+194+1=9130α=arccos(9130).ρ(S,TV)=1.5(4)3+294+1=1413;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}};


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