Question

The lengh of a line segment LM is 13 units .L is at( -2,4)and M is at (10,y). Find the positive value of y

Answer 1

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y = 9

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${\huge{\underline{\sf {\blue{Given :}}}}}$

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LM = 13 units

L = (-2 , 4)

M = (10 , y)

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Positive value of y

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${\huge{\underline{\sf {\blue{Formula \:\: used : }}}}}$

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$\sf \: for \: points \: ({x}_{1}{y}_{1}) \: and \: ({x}_{2}{y}_{2})$the distance between them is $\sf \sqrt{{({x}_{2} - {x}_{1})}^{2} + {({y}_{2} - {y}_{1})}^{2} }$

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$\sf \: {x}_{1} = - 2 ,\: {x}_{2} \: = \: 10, \: {y}_{1} \: = \: 4 ,\: {y}_{2} \: = \: y$

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$\sf \: Distance = \sqrt{ [{10 - ( - 2)]}^{2} + {(y - 4)}^{2} }$

↬ $\sf13 = \sqrt{ {(12)}^{2} + {y}^{2} - 8y + 16 }$

↬ $\sf ({13})^{2} = 144 + {y}^{2} - 8y + 16$

↬ $\sf169 = {y}^{2} - 8y + 160$

↬ $\sf {y}^{2} - 8y - 9 = 0$

↬ $\sf {y}^{2} - 9y - + y - 9 = 0$

↬ $\sf \: y(y - 9) + 1(y - 9) = 0$

↬ $\sf \: (y + 1)(y - 9) = 0$

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either,

y + 1 = 0

y = - 1

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or,

y - 9 = 0

y = 9

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We will ignore y = - 1 since the question requires only positive value of y.

Answer 2

Answer:

Step-by-step explanation:

D= $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

13=$\sqrt{(10-(-2))^2+(y-4)^2}$

$13^{2} =12^{2}+(y-4)^2\\169-144=(y-4)^2\\\sqrt{25} =\sqrt{(y-4)^2}\\5=y-4\\y=9$