question_answer

If \[A\,(5,2),\text{ }B\,(2,-2)\] and \[C(-2,t)\] are the vertices of a right angled triangle with \[\angle B=90{}^\circ \]. then find the value of t.

Answer:

Given, ABC are the vertices of a right angled triangle, then,

By Pythagoras theorem,

\[{{(AC)}^{2}}={{(BC)}^{2}}+{{(AB)}^{2}}\]                                 ?(i)

Now,                 \[{{(AC)}^{2}}={{(5+2)}^{2}}+{{(2-t)}^{2}}\]

\[=49+{{(2-t)}^{2}}\]

\[{{(BC)}^{2}}={{(2+2)}^{2}}+{{(-2-t)}^{2}}\]

\[=16+{{(t+2)}^{2}}\]

And                  \[{{(AB)}^{2}}={{(5-2)}^{2}}+{{(2+2)}^{2}}\]

\[=9+16=25\]

Putting these values in (i)

\[49+{{(2-t)}^{2}}=16+{{(t+2)}^{2}}+25\]

\[49+{{(2-t)}^{2}}=41+{{(t+2)}^{2}}\]

\[8={{(t+2)}^{2}}-{{(2-t)}^{2}}\]

\[8={{t}^{2}}+4+4t-4-{{t}^{2}}+4t\]

\[8=8t\]

\[\therefore t=1\]