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) \[49+{{(2-t)}^{2}}=16+{{(t+2)}^{2}}+25\] \[49+{{(2-t)}^{2}}=41+{{(t+2)}^{2}}\] \[8={{(t+2)}^{2}}-{{(2-t)}^{2}}\] \[8={{t}^{2}}+4+4t-4-{{t}^{2}}+4t\] \[8=8t\] \[\therefore t=1\]
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