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The point C(k,7) is equidistant from the points A(-5,1) and B(3,5). Find the value of k

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Let \(C(k,7) = (x_1,y_1), A(-5,1) = (x_2,y_2), B(3, 5) = (x_3, y_3)\)

Since points are equidistant, so

AC = BC

\(AC^2 = BC^2\)

Using distance formula,

\((x_3-x_1)^2 + (y_3-y_1)^2 = (x_2-x_1)^2+(y_2-y_1)^2\)

\(\Rightarrow (3-k)^2+(5-7)^2 = (-5-k)^2+(1-7)^2\)

\(\Rightarrow 9 - 6k + k^2 + (-2)^2 = 25 + 10k + k^2 + (-6)^2\)

\(\Rightarrow 9-6k+k^2+4=25+10k+k^2+36\)

\(\Rightarrow 13-6k=61+10k\)

\(\Rightarrow -6k-10k =61-13\)

\(\Rightarrow -16k = 48\)

\(\Rightarrow k = \cfrac{48}{-16}\)

\(\Rightarrow k = -3\)

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