Answer:
Ordinate of the point (y) = 3
Step-by-step explanation:
[tex]Let \:A(x,y) = (10,y) ,and \:P(2,-3)[/tex]
[tex] AP = 10\:(given)[/tex]
[tex] 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}}[/tex]
[tex]Now,PA = 10\\PA^{2}= 10^{2}\\(10-2)^{2}+[y-(-3)]^{2}=100[/tex]
[tex]\implies 8^{2}+(y+3)^{2}=100[/tex]
[tex]\implies 64+(y+3)^{2}=100[/tex]
[tex]\implies (y+3)^{2}=100-64=36[/tex]
[tex]\implies y+3 = \sqrt{36}[/tex]
[tex]\implies y+3 = 6[/tex]
[tex]\implies y = 6-3=3[/tex]
Therefore,
Ordinate of the point (y) = 3
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