find the value for y fir which fistance between the points p(2,-3) and q(10,y) is 10 units
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Answer:
√{(10-2) ²+[y-(-3)] ²}=10
squaring both sides
(10-2) ²+[y-(-3)]²=100
(8) ²+[y+3]²=100
64+y²+9+6y=100
y²+6y+73-100=0
y²+6y-27=0
y²+9y-3y-27=0
y(y+9) -3(y+9) =0
(y+9) (y-3) =0
If y+9=0, if y-3= 0
y=-9 y=3
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