+269 In triangle $STU$, let $M$ be the midpoint of $\overline{ST},$ and let $N$ be on $\overline{TU}$ such that $\overline{SN}$ is an altitude of triangle $STU$. If $ST = 13$, $SU = 14$, $TU = 4$, and $\overline{SN}$ and $\overline{UM}$ intersect at $X$, then what is $UX$?
+14933 What is UX?
\(\color{blue}S(0;0)\\\color{blue} T(13,0)\\ \color{blue}M(6.5, 0)\\f_S(x)=\pm \sqrt{14^2-x^2}\\ f_T(x)=\pm \sqrt{4^2-(x-13)^2}\\ x^2-14^2=(x-13)^2-4^2\\ x^2-14^2=x^2-26x+13^2-4^2\\ 26x=169-16+196\\ x_U=\dfrac{349}{26}\\ x_U=13.4\overline{230769}\\ y_U=3.9776\)
\(\color{blue}U(13.423,\ 3.978)\\ m_{TU}=\dfrac{y_U}{x_U-x_T}=\dfrac{3.9776}{13.42308-13}=9.4015\\ m_{NS}=-\dfrac{1}{m_{TU}}=-0.10637\)
\({\color{blue}f_{SN}(x)=}m_{NS}(x-x_S)+y_S=\color{blue}-0.10637x\)
\(m_{UM}=\dfrac{y_U}{x_U-x_M}=\dfrac{3.9776}{13.423-6.5}\\ m_{UM}=0.5387\)
\(f_{UM}(x)=m_{UM}(x-x_M)+y_M\\ f_{UM}(x)=0.5387x-0.5387\cdot 6.5\\ f_{UM}(x)=0.5387x-3.502\)
\(f_{UM}(x)=f_{SN}(x)\\ 0.5387x-3.5020=-0.10637x\\ x_X=\dfrac{3.5020}{0.5387+0.10637}\\ x_X=5.4285\\y_X=-0.57743\\ \color{blue}X(5.4285, -0.57743)\)
\(\overline{UX}=\sqrt{(y_U-y_X)^2+(x_U-x_X)^2}\\ \overline{UX}=\sqrt{(3.9776-(-0.57743))^2+(13.4231-5.4285)^2}\\ \color{blue}\overline{UX}=9.2012\)
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