A Line is of length 10 and one end is $(2, - 3)$ . If the abscissa of the other end is 10, then find its ordinate.

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Solution

Using distance formula:
$\sqrt{(10 - 2)^{2} + (y + 3)^{2}} = 10$
$\Rightarrow 64 + y^{2} + 9 + 6 y - 100 = 0$
$\Rightarrow y^{2} + 6 y - 27 = 0$
$\Rightarrow y = - 9, 3$

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Q. A line is of length 10 and one end is at the point (2, -3); if the abscissa of the other end be 10, prove that its ordinate must be 3 or -9.

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. Find its abscissa.

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A Line is of length 10 and one end is (2,−3). If the abscissa of the other end is 10, then find its ordinate.

Using distance formula:√(10−2)2+(y+3)2=10⇒64+y2+9+6y−100=0⇒y2+6y−27=0⇒y=−9,3

A Line is of length 10 and one end is $(2, - 3)$ . If the abscissa of the other end is 10, then find its ordinate.

Using distance formula:
$\sqrt{(10 - 2)^{2} + (y + 3)^{2}} = 10$
$\Rightarrow 64 + y^{2} + 9 + 6 y - 100 = 0$
$\Rightarrow y^{2} + 6 y - 27 = 0$
$\Rightarrow y = - 9, 3$