Find the point on the X axis which is equidistant from (2‚-5)and(-2‚9)
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let the point be p(x,y)
by section formula
p(x,y)={(mx2 - nx1)/2,(my1 - ny1)/2}
p(x,0)= {(-2-2)/2,(9+5)/2}
={-4/2,14/2}
={-2,7}
x=-2. ;. y=0
==x=-2
by section formula
p(x,y)={(mx2 - nx1)/2,(my1 - ny1)/2}
p(x,0)= {(-2-2)/2,(9+5)/2}
={-4/2,14/2}
={-2,7}
x=-2. ;. y=0
==x=-2
inumularicky4:
Why Y became 0
b-coz it is given in the que
that we have to find a point from x axis
There is both(-2,7) why you pick -2 as X
b-coz it is given in the question read the question carefully
not correct
any point which lies on x axis it has coordinate (x,0)
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