Question

Find all points having an x-coordinate of 4 whose distance from the point (-4,2) is 10. ​

Answer 1

Answer:

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

$\sqrt{ {8}^{2} + {(y - 2)}^{2} } = 10$

take square on both side.

8² + (y-2) ² = 100

(y-2) ² =100-64

(y-2) ² = 36

take root on both sides

y-2= + 6, -6

so y can be -8, -4

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Let (4, y) be such a point

So, we want to find y so that the distance from (-4,2) to (4,y) is 10

By the distance formula, √[(4-(-4))2+(y-2)2] = 10

Simplifying, we get: √[64+y2-4y+4] = 10

Square both sides to obtain: y2-4y+68 = 100

y2-4y-32 = 0

(y-8)(y+4) = 0

y = 8 or y = -4

NOTE: Since we squared both sides we need to check both values of y in the original equation to make sure that there are no extraneous solutions. Doing this, we see that both values of y satisfy the original equation.

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