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Abscissa
The ordinate of a point is 6 and the distance from origin to the point is 9. Find the abscissa of the
point.
Solution
We know that we can find the ordinate and the abscisa of a point, drawing perpendiculars from
the given point to the x and y axis.
We'll form a right angle triangle, whose hypotenuse is the distance from origin to the point and
cathetus is its abscisa.
We'll note the distance as r:
r = 9 units.
r^2 = 81 square units
We'll note the unknown abscisa as x and the ordinate as y.
y = 6 units
y^2 = 36 square units
We'll calculate x using Pythagorean Theorem:
r^2 = x^2 + y^2
x^2 = r^2 - y^2
x^2 = 81 - 36
x^2 = 45
x1 = sqrt 45
x1 = 3sqrt 5
x2 = - 3 sqrt 5
The ordinate of the point could be x = 3sqrt 5 or x = -3sqrt 5.

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AbscissaThe ordinate of a point is 6 and the distance from origin .pdf

  • 1. Abscissa The ordinate of a point is 6 and the distance from origin to the point is 9. Find the abscissa of the point. Solution We know that we can find the ordinate and the abscisa of a point, drawing perpendiculars from the given point to the x and y axis. We'll form a right angle triangle, whose hypotenuse is the distance from origin to the point and cathetus is its abscisa. We'll note the distance as r: r = 9 units. r^2 = 81 square units We'll note the unknown abscisa as x and the ordinate as y. y = 6 units y^2 = 36 square units We'll calculate x using Pythagorean Theorem: r^2 = x^2 + y^2 x^2 = r^2 - y^2 x^2 = 81 - 36 x^2 = 45 x1 = sqrt 45 x1 = 3sqrt 5 x2 = - 3 sqrt 5 The ordinate of the point could be x = 3sqrt 5 or x = -3sqrt 5.