Question

The length of a line-segments is 10. If one end is at (2, -3) and the abscissa of the second end is 10, show that its ordinate is either 3 or -9.

Answer 1

I have found this answer by using the distance formula from the chapter of std 10 named as coordinate geometry

Answer 2

Given: The length of a line-segments is 10. If one end is at (2, -3) and the abscissa of the second end is 10.

To Find: Show that its ordinate is either 3 or -9.

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Let A (2, -3) be the first end point and B (10, y) be the second end point.

✞Using distance formula:

${ \underline{ \boxed{ \bf {ab ={ \sqrt{{( { x_{2} - x_{1}) }^{2} } +{ ( { y_{2} - y_{1}) }^{2}}}}}}}} \:\red {\bigstar}$

$: \implies \sf10 = \sqrt{{ {(10 - 2)}^{2} } + ({y + 3)}^{2} } \\ \\ \\ : \implies \sf10 = \sqrt{ {8}^{2} + {y + 9 + 6y}^{2} } \\ \\ \\ \implies \sf10 = \sqrt{ 64 + {y}^{2} + 9 + 6y} \\ \\ \\ : \implies \sf64 + {y}^{2} + 9 + 6y = 100 \\ \\ \\ \sf : \implies {y}^{2} + 6y - 27 = 0 \\ \\ \\ \sf : \implies{y}^{2} + 9y - 3y - 27 = 0 \\ \\ \\ \sf : \implies y(y + 9) - 3(y + 9y) = 0 \\ \\ \\ \sf : \implies(y + 9) - (y - 3) = 0$

• Hence, y = -9 or y = 3.

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