question_answer

If \[A(4,3),B(-1,y)\] and \[C(3,4)\] are the vertices of a right triangle ABC, right-angled at A, then find the value of y.

Answer:

Given the triangle ABC, right angled at A.

Now,                 \[AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]

\[AB=\sqrt{{{(-1-4)}^{2}}+{{(y-3)}^{2}}}\]

\[AB=\sqrt{{{(-5)}^{2}}+{{(y-3)}^{2}}}\]

\[AB=\sqrt{25+{{(y-3)}^{2}}}\]

\[AB=\sqrt{25+{{y}^{2}}+9-64}\]

\[AB=\sqrt{34+{{y}^{2}}-6y}\]

\[BC=\sqrt{{{(3-(-1))}^{2}}+{{(4-y)}^{2}}}\]

\[BC=\sqrt{{{(4)}^{2}}+{{(4-y)}^{2}}}\]

\[BC=\sqrt{16+16+{{y}^{2}}-8y}\]

\[BC=\sqrt{32+{{y}^{2}}-8y}\]

And                  \[AC=\sqrt{{{(3-4)}^{2}}+{{(4-3)}^{2}}}\]

\[AC=\sqrt{{{(-1)}^{2}}+{{(1)}^{2}}}\]

\[AC=\sqrt{1+1}\]

\[AC=\sqrt{2}\] units

Given, \[\Delta \text{ }ABC\] is a right angled triangle

So, by Pythagoras theorem

\[B{{C}^{2}}=A{{C}^{2}}+A{{B}^{2}}\]

\[{{(\sqrt{32+{{y}^{2}}-8y})}^{2}}={{(\sqrt{2})}^{2}}+{{(\sqrt{34+{{y}^{2}}-6y})}^{2}}\]

\[32+{{y}^{2}}-8y=2+34+{{y}^{2}}-6y\]

\[-2y=4\]

\[y=-2\]

Hence, the value of y is \[-2\].