find the value of y for which the distance between the points P (7,-5) and Q(15,y) is 10 units
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distance formula= √ (x2 -x1)^2 +( y2 - y1)^2
10 = √ (15-7)^2 + (y-(-5)^2
on squaring both sides
100 = (8)^2 + ( y+5)^2
100 = 64 + y^2+25 +10y
y^2+10y -11 = 0
y^2-1y+11y -11 =0
y(y -1)+11(y -1) =0
(y -1)(y+11) =0
y = 1 or -11
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