Question

One end of a line of length 10 units is the point (3,-2).if the ordinate of the other end is 10 ,what is its abscissa?

Answer 1

You can use distance formula for solving.
(x2-x1)^2+(y2-y1)^2
(3-x)^2+(-2-10)^2=10
Solve this. You will get the answer.

Answer 2

Given:

One end of a line of length 10 units is the point (3,-2).if the ordinate of the other end is 10.

Ordinate is y value of coordinate.

Calculation:

Let abscissa of other end is x

Other end coordinate, P(x,10)

One end coordinate, Q(3,-2)

PQ=10

Formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$PQ=\sqrt{(x-3)^2+(10+2)^2}$

$10^2= (x-3)^2+(10+2)^2$

$100=(x-3)^2+144$

$(x-3)^2=-44$

Square never gives negative value.

Solution does not exist.

No such value of which satisfy given condition.

No solution