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# Math Help

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A line segment begins at (2, 5). It is 10 units long and ends at the point (-6, y) where \$y > 0. What is the value of y?

Jun 27, 2018

#1
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By the distance formula...

distance between  (2, 5)  and  (-6, y)   =   $$\sqrt{(-6-2)^2+(y-5)^2}$$

The problem tells us that the distance between  (2, 5)  and  (-6, y)  is  10 units, so.....

$$\sqrt{(-6-2)^2+(y-5)^2}=10$$

Square both sides of the equation.

(-6 - 2)2 + (y - 5)2  =  100

Multiply out each exponent on the left side.

( -8 )2 + (y - 5)(y - 5)  =  100

64 + y2 - 10y + 25  =  100

Combine  64  and  25  to get  89

y2 - 10y + 89  =  100

Subtract  100  from both sides of the equation.

y2 - 10y - 11  =  0

Factor the left side.

(y - 11)(y + 1)  =  0

Set each factor equal to zero and solve for  y .

y - 11  =   0       or       y + 1  =  0

y  =  11                      y  =  -1

Since  y > 0  ,  the solution must be  y = 11 Jun 27, 2018

#1
+2

By the distance formula...

distance between  (2, 5)  and  (-6, y)   =   $$\sqrt{(-6-2)^2+(y-5)^2}$$

The problem tells us that the distance between  (2, 5)  and  (-6, y)  is  10 units, so.....

$$\sqrt{(-6-2)^2+(y-5)^2}=10$$

Square both sides of the equation.

(-6 - 2)2 + (y - 5)2  =  100

Multiply out each exponent on the left side.

( -8 )2 + (y - 5)(y - 5)  =  100

64 + y2 - 10y + 25  =  100

Combine  64  and  25  to get  89

y2 - 10y + 89  =  100

Subtract  100  from both sides of the equation.

y2 - 10y - 11  =  0

Factor the left side.

(y - 11)(y + 1)  =  0

Set each factor equal to zero and solve for  y .

y - 11  =   0       or       y + 1  =  0

y  =  11                      y  =  -1

Since  y > 0  ,  the solution must be  y = 11 hectictar Jun 27, 2018