Question

The length of a line segment is of 10 units and the coordinates of one end-point are (2,-3). If the abscissa of the other end is 10, find the ordinate of the other end.

Answer 1

Answer:

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}$

Let ordinatre of the point be y

So,

√ { (2-10)^{2} + (-3-y)^{2} } = 10(2−10)2+(−3−y)2=10

or, 64 + (y+3)(y+3) = 100

or, (y+3)(y+3) = 36

or, y+3 = 6,-6

or, y = 6-3, -6-3 = 3,-9 (proved)

Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

Let ordinatre of the point be y

So,

\sqrt{ (2-10)^{2} + (-3-y)^{2} } = 10

or, 64 + (y+3)(y+3) = 100

or, (y+3)(y+3) = 36

or, y+3 = 6,-6

or, y = 6-3, -6-3 = 3,-9 (proved)