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 \[-2\].
You need to login to perform this action.
You will be redirected in
3 sec