(11.) If the distance between two points (x, 7) and (1, 15) is 10, find the value of t.
Answers
Answer:Distance cannot be negative so x=7
Step-by-step explanation:
By using distance formula,
√(y2-y1)²+(x2-x1)²=10
√(15-7)²+(1-x)²=10
64+1+x²-2x=100
x²-2x-35=0
x²-7x+5x-35=0
x(x-7)5(x-7)=0
Therefore,x=7 or -5
Distance cannot be negative so x=7
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Answer:
We have to find the value of x not t .
By Distance formula,
√((x_2-x_1)²+(y_2-y_1)²) = 10
Square on both sides,
(x_2-x_1)²+(y_2-y_1)² = 100
(1 - x)² + (15 - 7)² = 100
( 1 - 2x + x² ) + ( 8 )² = 100
( 1 - 2x + x²) + 64 = 100
65 - 2x + x² = 100
x² - 2x = 100 - 65
x² - 2x = 35
x² - 2x - 35 = 0
Now split the middle terms,
x² - 7x+5x - 35 = 0
x ( x - 7 ) + 5 ( x - 7 ) = 0
( x - 7 ) ( x + 5 )
x = 7 or x = -5
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