Question

The length of a line segment $AB$ is $10$ units. If the coordinates of one extremity are $(2,-3)$ and the abscissa of the other extremity is $10$ then the sum of all possible values of the ordinate of the other extremity is

• A. $3$
• B. $-4$
• C. $12$
• D. $-6$

Answer 1

The correct option is D $-6$

Let the ordinate be $y$

$\therefore$ Point $1\left(A\right)\equiv (2,-3)$

$\therefore$ Point $2\left(B\right)\equiv (10,4)$

$\therefore d(A,B)=\sqrt{(2-10{)}^2+(-3-4{)}^2}$

$\therefore \sqrt{64+(3+y{)}^2}=10$

$\therefore 36=(3+y{)}^2$

$\therefore 3+y=6$ OR $3+y=-6$

$\therefore y=3$ OR $y=-9$

$\therefore$ Sum $=3+(-9)=-6$