Question

If the point (x, y) is at equal distance from the x-axis and y-axis. Then, the relationship between x and y can be 1.x = 2y 2.x = 3y 3.2x + y = 0 4.x + y = 0

Answer 1

Therefore the relationship between x and y can be x + y = 0.

Given : If the point (x, y) is at equal distance from the x-axis and y-axis.

To find : The relationship between x and y can be .

1. x = 2y
2. x = 3y
3. 2x + y = 0
4. x + y = 0

solution : a/c to question, point (x, y) is at equal distance from the x-axis and y-axis.

point on x - axis = (x, 0)

point on y - axis = (0, y)

using distance formula,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

distance between (x,y) and (x, 0) = distance between (x,y) and (0, y)

⇒$\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2}$

⇒|y| = |x|

⇒y = ± x

⇒x + y = 0 or x - y = 0

Therefore the relationship between x and y can be x + y = 0.

Answer 2

Answer:

Therefore the relationship between x and y can be x + y = 0.

Given : If the point (x, y) is at equal distance from the x-axis and y-axis.

To find : The relationship between x and y can be .

x = 2y

x = 3y

2x + y = 0

x + y = 0

solution : a/c to question, point (x, y) is at equal distance from the x-axis and y-axis.

point on x - axis = (x, 0)

point on y - axis = (0, y)

using distance formula,

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

distance between (x,y) and (x, 0) = distance between (x,y) and (0, y)

⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2}

(x−x)

2

+(y−0)

2

=

(x−0)

2

+(y−y)

2

⇒|y| = |x|

⇒y = ± x

⇒x + y = 0 or x - y = 0

Therefore the relationship between x and y can be x + y = 0.

Step-by-step explanation:

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