Question

find the ordinate of a point whose abscissa is 10 and which is at a distance of 10 units from the point P(2,-3).

Answer 1

Answer:

Ordinate of the point (y) = 3

Step-by-step explanation:

$Let \:A(x,y) = (10,y) ,and \:P(2,-3)$

$AP = 10\:(given)$

$By\: distance \: formula:\\If \: A(x_{1},y_{1})\:and\:B(x_{2},y_{2})\:then \\Distance (AB) =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$Now,PA = 10\\PA^{2}= 10^{2}\\(10-2)^{2}+[y-(-3)]^{2}=100$

$\implies 8^{2}+(y+3)^{2}=100$

$\implies 64+(y+3)^{2}=100$

$\implies (y+3)^{2}=100-64=36$

$\implies y+3 = \sqrt{36}$

$\implies y+3 = 6$

$\implies y = 6-3=3$

Therefore,

Ordinate of the point (y) = 3

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Answer 2

Step-by-step explanation:

Hope this helped you....................

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