Maharashtra BoardEnglish MediumSTD 10MathsP-2 Co-ordinate Geometry3 Marks
Question
Find the point on the X-axis which is equidistant from A(-3, 4) and B(1, -4).
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Answer
A point in the x = axis is of the form (a,0)
Distance d between two points(a,b) and (c,d)is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}$
Distance between $(-3,4)$ and $(a, 0)=$
$D=\sqrt{(-3-a)^2+(0-4)^2} $
$D \sqrt{(3+a)^2+16}$
Distance between $(1,-4)$ and $(a, 0)$
$D=\sqrt{(1-a)^2+(0-(-4))^2} $
$D=\sqrt{(1-a)^2+16}$
As the two points are equidistant from the point (a.0)
$\sqrt{(1-a)^2+16}=\sqrt{(3+a)^2+16}$
Squaring both sides, we get
$(1-a)^2+16=(3+a)^2+16$
$1+a^2-2 a=9+a^2+6 a$
$8 a=-8$
$a=-1$
Hence the point is $(-1,0)$
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