find the ordinateof a point whose abscissa is 10 and which is at a distance of 10 unit from the point P(2,-3).
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a unknown point which abcissa is 10 means x=10 let ordinate (y) of this point
use distance formula ,
10=root {(10-2)^2+(y-(-3))^2}
squaring borh side .
100=8^2+(y+3)^2
36=(y+3)^2
(y+3)^2=(6)^2
y+3=+_6
y=-3+6,-3-6
=3,-9
hence points are (10,3) and (10,-9)
use distance formula ,
10=root {(10-2)^2+(y-(-3))^2}
squaring borh side .
100=8^2+(y+3)^2
36=(y+3)^2
(y+3)^2=(6)^2
y+3=+_6
y=-3+6,-3-6
=3,-9
hence points are (10,3) and (10,-9)
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