If the distance between the point (×,2) and (3,-6) is 10 units. What

1.	If the distance between the point (×,2) and (3,-6) is 10 units. What is the positive value of ×
Answer» (X2-X1)^2+(Y2-Y1)^2 10^2=(3-X)^2+(-6-2)^2 100= (3^2+X^2-2×3X)+64 100-64=9+X^2-6X 36-9=X^2-6X X^2-6X-27=0 X= 9 or -3 hence, POSITION vale of X is 9 or -3

If the distance between the point (×,2) and (3,-6) is 10 units. What is the positive value of ×

Answer»

(X2-X1)^2+(Y2-Y1)^2

10^2=(3-X)^2+(-6-2)^2

100= (3^2+X^2-2×3X)+64

100-64=9+X^2-6X

36-9=X^2-6X

X^2-6X-27=0

X= 9 or -3

hence, POSITION vale of X is 9 or -3

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If the distance between the point (×,2) and (3,-6) is 10 units. What is the positive value of ×

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