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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?