Question

A line is of length 13 units and one end is at (5,3).If the abscissa of the other end is 17.Then its ordinate is

Answer 1

by solving this we get quadratic eq. as x^2-10x-68..by this we get two values of ordinates..

Answer 2

$total \: length \: = 13units \\ \\ one \: coordinate = (5 \: and \: 3) \\ \\ another \: point \: = (17and \: y2) \\ \\ distance \: between \: these \: points \: = total \: length \\ \\ = > \sqrt{(x2 - x1) {}^{2} + (y2 - y1) {}^{2} } = 13 \\ \\ = > \sqrt{(17 - 5) {}^{2} + (y2 - 3) {}^{2} } = 13 \\ \\ = > \sqrt{144 + (y2 - 3) {}^{2} } = 13 \\ \\ squaring \: on \: both \: sides \\ \\ = > 144 + (y2 - 3) {}^{2} = 169 \\ \\ = > 144 + (y2) {}^{2} + 9 - 6y2 = 169 \\ \\ = > (y2) {}^{2} - 6y2 - 16 = 0 \\ \\ = > (y2) {}^{2} - 8y2 + 2y2 - 16 = 0 \\ \\ = > y2(y2 - 8) + 2(y2 - 8) = 0 \\ \\ = > (y2 - 8)(y2 + 2) = 0 \\ \\ = > y2 = 8 \: \: (or) \: \: y2 = - 2$

∴ The ordinate is 8 (or) -2