Question

Find the point on y-axis which are at a distance of 10 units from the point (6,4)
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Answer 1

Answer:

using distance formula,

S=$\sqrt{(x2-x1)^{2} +(y2-y1)^{2} }$

Let (0,y)be the point

S=$\sqrt{(x2-x1)^{2} +(y2-y1)^{2} }$

10=$\sqrt{(6-0)^{2} +(4-y)^{2} }$

10=$\sqrt{6^{2}+(4-y)^{2} }$

10=$\sqrt{36+16+y^{2} }$

10=6+4+y

10=10+y

y=0

Point on y-axis=0

Step-by-step explanation: