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Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, -4) |
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Answer» Given: Points are A(1, 5, 7), B(5, 1, -4) To find: the point on z-axis which is equidistant from the points As we know x = 0 and y = 0 on z-axis Let R(0, 0, z) any point on z-axis According to the question: RA = RB ⇒ RA2 = RB2 Formula used: The distance between any two points (a, b, c) and (m, n, o) is given by, \(\sqrt{(a-m)^2+(b-n)^2+(c-0)^2}\) Therefore, Distance between R(0, 0, z) and A(1, 5, 7) is RA, = \(\sqrt{(0-1)^2+(0-5)^2+(z-7)^2}\) = \(\sqrt{1+25+(z-7)^2}\) = \(\sqrt{26+(z-7)^2}\) The distance between P(x, y, 0) and B(2, 1, 2) is PB, = \(\sqrt{(0-5)^2+(0-1)^2+(z-(-4)^2}\) = \(\sqrt{(z+4)^2+25+1}\) = \(\sqrt{(z+4)^2+26}\) As RA2 = RB2 26+ (z – 7)2 = (z + 4)2 + 26 ⇒ z2+ 49 – 14z + 26 = z2+ 16 + 8z + 26 ⇒ 49 – 14z = 16 + 8z ⇒ 49 – 16 = 14z + 8z ⇒ 22z = 33 ⇒ z = \(\frac{33}{22}\) ⇒ z = \(\frac{3}{2}\) Hence point \(\Big(0,0,\frac{3}{2}\Big)\) on z-axis is equidistant from (1, 5, 7) and (5, 1, -4) |
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