Question

If A (5,2),B (2,-2) and C (-2,t) are the vertices of a right angled triangle with B=90°,then find the value of t.

Answer 1

Answer:

Step-by-step explanation:

HOPE U WILL UNDERSTAND THIS❤

Answer 2

Answer:

$\red { Value \: of \: t }\green {= 1}$

Step-by-step explanation:

$Let \: A(5,2) = (x_{1},y_{1}),\\B(2,-2) = (x_{2},y_{2}),and \:C(-2,t) = (x_{3},y_{3})\:are \\veritcies \:of \:right\:angled\:triangle$

$\angle B = 90\degree$

$\blue { \underline { By \:using \: distance \: formula:}}$

$AB^{2} = ( x_{2}-x_{1})^{2} + ( y_{2}-y_{1})^{2}\\=(2-5)^{2}+(-2-2)^{2}\\= (-3)^{2}+(-4)^{2}\\= 9 + 16 \\= 25$

$BC^{2} = ( x_{3}-x_{2})^{2} + ( y_{3}-y_{2})^{2}\\=(-2-2)^{2}+(t+2)^{2}\\= (-4)^{2}+(t+2)^{2}\\= 16 + (t+2)^{2}$

$CA^{2} = ( x_{3}-x_{1})^{2} + ( y_{3}-y_{1})^{2}\\=(-2-5)^{2}+(t-2)^{2}\\= (-7)^{2}+(t-2)^{2}\\= 49 + (t-2)^{2}$

$\blue { \underline { By \: Phythagorean \: theorem:}}$

$CA^{2} = AB^{2} + BC^{2}$

$49 + (t-2)^{2} = 25 + 16 + (t+2)^{2}$

$\implies 49 - 41 = (t+2)^{2} - (t-2)^{2}$

$\implies 8 = 4\times t \times 2$

$\boxed { \pink {(a+b)^{2} - (a-b)^{2} = 4ab }}$

$\implies 8 = 8t$

$\implies \frac{8}{8} = t$

$\implies t = 1$

Therefore.,

$\red { Value \: of \: t }\green {= 1}$

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