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
2⋅2+3⋅5−1=18≠0⇒(2,5)∉VS:2x+3y−1=0⇒T=T(2,5)ST:y=13x+b,T∈ST:y=13x+b⇒b=133S={2x+3y−1=0y=13x+133⇒S=S(−4,3).TV:y−5=−1−23(x−2)=32(x−2).cos(α)=13⋅32+(−1)⋅(−1)19+1⋅94+1=9130⇒α=arccos(9130).ρ(S,TV)=∣1.5⋅(−4)−3+2∣94+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}};2⋅2+3⋅5−1=18=0⇒(2,5)∈VS:2x+3y−1=0⇒T=T(2,5)ST:y=31x+b,T∈ST:y=31x+b⇒b=313S={2x+3y−1=0y=31x+313⇒S=S(−4,3).TV:y−5=3−2−1(x−2)=23(x−2).cos(α)=91+1⋅49+131⋅23+(−1)⋅(−1)=1309⇒α=arccos(1309).ρ(S,TV)=49+1∣1.5⋅(−4)−3+2∣=1314;
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