Question

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find the co-ordinates of the points on y axis which are at the distance of 10 units from the point (6,6)​​

Answer 1

Step-by-step explanation:

Given :-

The point (6,6)

To find :-

Find the co-ordinates of the points on y axis which are at the distance of 10 units from the point (6,6) ?

Solution :-

Given point = (6,6)

We know that

The equation of y-axis is x = 0

So , the required point be P (0,y)

Let (x1, y1) = A(6,6) => x1 = 6 and y1 = 6

Let (x2, y2) = P(0,y) => x2 = 0 and y2 = y

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

The distance between two points A and P

=> AP = √[(0-6)²+(y-6)²]

=> AP = √[(-6)²+(y-6)²]

=> AP = √(36+y²-12y+36)

=> AP = √(y²-12y+72) units

According to the given problem

The distance between the two points = 10 units

=> √(y²-12y+72) = 10

On squaring both sides then

=> [√(y²-12y+72)]² = 10²

=> y²-12y+72 = 100

=> y²-12y+72-100 = 0

=> y²-12y-28 = 0

=> y²+2y-14y-28 = 0

=> y(y+2)-14(y+2) = 0

=> (y+2)(y-14) = 0

=> y+2 = 0 or y-14 = 0

=> y = -2 or y = 14

Therefore y = -2 or 14

The point = (0,-2) or (0,14)

Answer :-

The required point for the given problem is (0,-2) or (0,14)

Used formulae:-

→ The equation of y-axis is x = 0

→The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

Answer 2

Answer :—

• The required point will be (0,-2 ) or (0,14)

Solution :—

$\implies{6 - 0}^{2} + \: { \:6 - y}^{2} = \: 100 \\ \\ = {36 + (6 - y)}^{2} = \: 100 \\ = \: {(6 - y)}^{2} = \: 100 - 36 \: = 64 \\ = Therefore \: (6 - y) \: = \: \sqrt{64 } \\ = + 8 \: or \: - 8 \\ If \: 6 \: - \: y \: = 8 \: then \: y \: = \: 8 - 6 \: = 2 \\ Therefore \: y \: = \: - 2 \\ If \: 6 - y \: = \: - 8 \: then \: \: y \: = - 8 - 6 \: = - 14$

$\longrightarrow$ The required point will be (0,-2 ) or (0,14) .