The abscisa of a point is -3 and its distance from the origin is 5. What is its ordinate?
Let us have the point P whose absissa is -3 and the distance fro the origin is 5. We required to find the ordinate of the point P.
If ( x,y) are the absissa and oridinates of the point P , then its distance from the origin is given by d^2 = (x-0)2+(y-0)^2.
Therefore d ^2 = x^2+y^2.
Given d= 5, x = -3, we have to determine y.
Substituting in the formula , d- 5, and x = -3 in the formula d^2= x^2+y^2 , we get:
5^2 +(-3)^2 +y^2.
Therefore 25 = 9 +y^2.
y ^2 = 25-9.
y ^2 = 16.
y = sqrt(16).
y = 4 or y = y = -4.
Therefore for the given absissa o -3 , there two ordinates 4 and -4 for which the distance from the origin is 5.
So (-3 , -4) and (-3 4 ) are the two points which satisfy the conditions of x and y coordinates with a distance of 5 units from the origin.
We know that we can find the abscisa and the ordinate 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 = 5 units.
r^2 = 25 square units
We'll note the abscisa as x:
x = -3 units
x^2 = 9 square units
We'll calculate y using Pythagorean Theorem:
r^2 = x^2 + y^2
y^2 = r^2 - x^2
y^2 = 25 - 9
y^2 = 16
y1 = 4
y2 = -4
The ordinate of the point could be y = 4 or y = -4.