What is the perpendicular distance of point P (- 2 3 from the axis?

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The distance of the point P(2, 3) from the x-axis is ______.

The distance of the point P(2, 3) from the x-axis is 3.

Explanation:

We know that,

(x, y) is a point on the Cartesian plane in first quadrant.

Then,

x = Perpendicular distance from Y-axis and

y = Perpendicular distance from X-axis

Therefore, the perpendicular distance from X-axis = y coordinate = 3

Concept: Distance Formula

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Page 2

The distance between the points A(0, 6) and B(0, –2) is ______.

The distance between the points A(0, 6) and B(0, –2) is 8.

Explanation:

Distance formula: d2 = (x2 – x1)2 + (y2 – y1)2

We have,

x1 = 0, x2 = 0

y1 = 6, y2 = – 2

d2 = (0 – 0)2 + (– 2 – 6)2

d= `sqrt((0)^2 + (- 8)^2)`

d = `sqrt(64)`

d = 8 units

Therefore, the distance between A(0, 6) and B(0, 2) is 8

Concept: Distance Formula

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Page 3

The distance of the point P(–6, 8) from the origin is ______.

The distance of the point P(–6, 8) from the origin is 10.

Explanation:

Distance formula: d2 = (x2 – x1)2 + (y2 – y1)2

We have;

x1 = – 6, x2 = 0

y2 = 8, y2 = 0

d2 = [0 – ( – 6)]2 + [0 – 8]2

d= `sqrt((0 - (-6))^2 + (0 - 8)^2`

d= `sqrt((6)^2 + (-8)^2)`

d = `sqrt(36 + 64)`

d = `sqrt(100)`

d = 10

Therefore, the distance between P(– 6, 8) and origin O(0, 0) is 10

Concept: Distance Formula

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Page 4

The distance between the points (0, 5) and (–5, 0) is ______.

The distance between the points (0, 5) and (–5, 0) is `5sqrt(2)`.

Explanation:

Distance formula: d2 = (x2 – x1)2 + (y2 – y1)2

We have;

x1 = 0, x2 = – 5

y2 = 5, y2 = 0

d2 = (( – 5) – 0)2 + (0 – 5)2

d= `sqrt((-5 - 0)^2 + (0 - 5)^2`

d= `sqrt((-5)^2 + (-5)^2)`

d = `sqrt(25 + 25)`

d= `sqrt(50) = 5sqrt(2)`

So the distance between (0, 5) and (– 5, 0) = `5sqrt(2)`

Concept: Distance Formula

  Is there an error in this question or solution?