The length of a line-segment is 10. If one end is at (2, -3) and the abscissa of the second end is 10, then find the number of values of ordinate the second end can take.

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Solution

Let $(2, - 3)$ be the point A Let the ordinate of the second end B be y Then its coordinates will be $(10, y)$
Given, $A B = \sqrt{{(10 - 2)}^{2} + {(y + 3)}^{2}} = 10$
$\sqrt{64 + y^{2} + 6 y + 9} = 10$
$y^{2} + 6 y + 73 = 100$
$y^{2} + 6 y - 27 = 0$
$(y + 9) (y - 3) = 0$
Therefore $y = - 9, y = 3$
i.e. the ordinate is $3$ or $- 9$

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The length of a line-segment is 10. If one end is at (2, -3) and the abscissa of the second end is 10, then find the number of values of ordinate the second end can take.

Let (2,−3) be the point A Let the ordinate of the second end B be y Then its coordinates will be (10,y)Given, AB=√(10−2)2+(y+3)2=10√64+y2+6y+9=10y2+6y+73=100 y2+6y−27=0(y+9)(y−3)=0Therefore y=−9 ,y=3i.e. the ordinate is 3 or −9