# The ordinate of a point is 6 and the distance from origin to the point is 9. Find the abscissa of the point.

*print*Print*list*Cite

### 3 Answers

Let the abscissa of the point be x. Therefore the point is (x, 6)

Now the distance of (x, 6) from the origin is given to be 9.

We also know that the distance between (x1, y1) and (x2, y2) is given by sqrt [ (x2- x1)^2 + (y2 - y1)^2]

Therefore here we have 9 = sqrt [ ( x - 0)^2 + ( 6- 0)^2]

=> 9= sqrt ( x^2 + 36)

=> 81 = x^2 + 36

=> x^2 = 45

=> x = sqrt 45 = 3 sqrt 5

or x = -sqrt 45 = -3 sqrt 5

**Therefore the abscissa of the point is 3 sqrt 5 or -3 sqrt 5**

We know that the distance d between the two points P(x1,y1) and Q(x2, y2) is giben by

d ^2 = PQ^2 = (x2-x1)^2 +(y2-y1)^2.

The given point A has the ordinate 6, d = 9, the abscissa is to be determined. Let the abscissa be x. Therefore the point A is A(x, 6).

So the distance of A(x, 9) from the origin O (0, 0) is given by:

OA^2 = (x-0)^2+(6-0)^2 = 9 ...(1)

We solve the eq (1) for x to get the absissa:

x^2 = 9^2- 6^2

x^2 = 81-36 = 45.

So x = sqrt45 = 3sqrt5 , or x= -sqrt45= -3sqrt5

Therefore the abssisa of A is x = -3sqrt5 or x= - 3sqrt5.

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. **