+0  
 
+1
409
2
avatar+836 

Let A,B  be the points on the coordinate plane with coordinates  \((t-4,-1)\) and\((-2,t+3)\)   , respectively. The square of the distance between the midpoint of \(\overline{AB}\)  and an endpoint of \(\overline{AB}\)  is equal to \(t^2/2\)  . What is the value of \(t\) ?

 Jul 27, 2017
 #1
avatar+7347 
+3

midpoint of AB  \(=\,(\frac{t-4+-2}{2}\,,\,\frac{-1+t+3}{2})\,=\,(\frac{t-6}{2}\,,\,\frac{t+2}{2})\)

 

 

an endpoint of AB  \(=\,(t-4,-1)\)

 

distance between midpoint and an endpoint  \(=\,\ \,\sqrt{(t-4-\frac{t-6}{2})^2+(-1-\frac{t+2}{2})^2}\)

 

 

\(\begin{array} \ {(t-4-\frac{t-6}{2})^2+(-1-\frac{t+2}{2})^2}\,&=&\,\frac{t^2}{2} \\~\\ {(\frac{2t}{2}-\frac{8}{2}-\frac{t-6}{2})^2+(-\frac{2}{2}-\frac{t+2}{2})^2}\,&=&\,\frac{t^2}{2} \\~\\ {(\frac{2t-8-t+6}{2})^2+(\frac{-2-t-2}{2})^2}\,&=&\,\frac{t^2}{2} \\~\\ {(\frac{t-2}{2})^2+(\frac{-t-4}{2})^2}\,&=&\,\frac{t^2}{2} \\~\\ {(\frac{t-2}{2})(\frac{t-2}2)+(\frac{-t-4}{2})(\frac{-t-4}{2})}\,&=&\,\frac{t^2}{2} \\~\\ {\frac{t^2-4t+4}{4}+\frac{t^2+8t+16}{4}} \,&=&\,\frac{t^2}{2} \\~\\ {\frac{2t^2+4t+20}{4}}\,&=&\,\frac{t^2}{2} \\~\\ 2t^2+4t+20 \,&=&\,2t^2 \\~\\ 4t+20\,&=&\,0 \\~\\ 4t\,&=&\,-20 \\~\\ t\,&=&\,-5 \end{array}\)

.
 Jul 27, 2017
edited by hectictar  Jul 27, 2017
 #2
avatar+20847 
+2

Let A,B  be the points on the coordinate plane with coordinates \(\tbinom{t-4}{-1}\)  and \(\tbinom{-2}{t+3}\), respectively.
The square of the distance between the midpoint of  \( \overline{AB}\) and an endpoint of   \(\overline{AB}\) is equal to  \(\tfrac{t^2}{2}\).
What is the value of \(t\) ?

 

\(\begin{array}{|rcll|} \hline \left ( \frac{ \overline{AB} } {2} \right)^2 &=& \frac{t^2}{2} \\ \frac{ \overline{AB}^2 } {4} &=& \frac{t^2}{2} \\ \mathbf{ \overline{AB}^2 } & \mathbf{=} & \mathbf{2 t^2 } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \overline{AB}^2 &=& [~ (t-4)-(-2) ~ ]^2 + [~ (-1) - (t+3) ~ ]^2 \\ \overline{AB}^2 &=& (~ t-4+2 ~)^2 + (~ -1 - t -3 ~ )^2 \\ \overline{AB}^2 &=& (~ t-2 ~)^2 + (~ - t -4 ~ )^2 \\ \overline{AB}^2 &=& (~ t-2 ~)^2 + [~ - (t +4) ~ ]^2 \\ \overline{AB}^2 &=& (~ t-2 ~)^2 + (~ t +4 ~)^2 \quad & | \quad \overline{AB}^2 = 2t^2\\ 2t^2 &=& (~ t-2 ~)^2 + (~ t +4 ~)^2 \\ 2t^2 &=& t^2 -4t + 4 + t^2 + 8t + 16 \\ 0 &=& -4t + 4 + 8t + 16 \\ 0 &=& 4t + 20 \\ 4t &= -20 \\ t &=& -\frac{20}{4} \\ \mathbf{ t } & \mathbf{=} & \mathbf{-5} \\ \hline \end{array}\)

 

 

laugh

 Jul 27, 2017
edited by heureka  Jul 27, 2017

9 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.