# The abscisa of a point is -3 and its distance from the origin is 5. What is its ordinate?

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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. **