Evaluate : 62-9Cb-1) if b= 1.1
[Hint : Put b = 1.1 =11/10]
Answers
Step-by-step explanation:
Answer:-
Given:-
Distance between two points S(- 2 , 3) & T(3 , y) = 13 units.
We know that,
Distance between two points (x₁ , y₁) & (x₂ , y₂) is
\red{\sf \: \sqrt{ {(x_2 - x_1)}^{2} + ( {y_2 - y_1)}^{2} }}
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Let,
x₁ = - 2
y₁ = 3
x₂ = 3
y₂ = y.
So,
According to the question;
\implies \sf \: \sqrt{ {(3 - ( - 2))}^{2} + {(y - 3)}^{2} } = 13⟹
(3−(−2))
2
+(y−3)
2
=13
On squaring both sides we get,
\begin{gathered} \implies \sf \: \bigg( \sqrt{ {(3 + 2)}^{2} + {(y - 3)}^{2} } \bigg) ^{2} = 13^{2} \\ \\ \\ \implies \sf \: {5}^{2} + {(y - 3)}^{2} = 169\end{gathered}
⟹(
(3+2)
2
+(y−3)
2
)
2
=13
2
⟹5
2
+(y−3)
2
=169
Using (a - b)² = a² + b² - 2ab we get,
\begin{gathered} \: \implies \sf \:25 + {y}^{2} + 9 - 6y = 169 \\ \\ \\ \implies \sf \: {y}^{2} - 6y + 34 - 169 = 0 \\ \\ \\ \implies \sf \: {y}^{2} - 6y - 135 = 0 \\ \\ \\ \implies \sf \: {y}^{2} + 9y - 15y - 136 = 0 \\ \\ \\ \implies \sf \:y(y + 9) - 15(y + 9) = 0 \\ \\ \\ \implies \sf \:(y + 9)(y - 15) = 0 \\ \\ \\ \implies \boxed{ \sf \:y = - 9 \: ,\: 15}\end{gathered}
⟹25+y
2
+9−6y=169
⟹y
2
−6y+34−169=0
⟹y
2
−6y−135=0
⟹y
2
+9y−15y−136=0
⟹y(y+9)−15(y+9)=0
⟹(y+9)(y−15)=0
⟹
y=−9,15
∴ The values of y are - 9 & 15 to satisfy the given condition.
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Step-by-step explanation:
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