Question

If point \(B(0, 1)\) is equidistant from points\( A(5, -3) \)and \(C(x, 6)\), then find the values of \( x\)

• A. \(\pm2\)
• B. \(\pm8\)
• C. \(\pm4\)
• D. \(\pm5\)

Answer 1

The Correct Option is C

Given that point \( B(0, 1) \) is equidistant from points \( A(5, -3) \) and \( C(x, 6) \), we need to find the values of \( x \).

To determine \( x \), use the distance formula. The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Apply the distance formula to find distances \( AB \) and \( BC \):

\[AB = \sqrt{(5-0)^2+(-3-1)^2} = \sqrt{5^2+(-4)^2} = \sqrt{25+16} = \sqrt{41}\]

\[BC = \sqrt{(x-0)^2+(6-1)^2} = \sqrt{x^2+5^2} = \sqrt{x^2+25}\]

Since \( B \) is equidistant from \( A \) and \( C \), the distances are equal:

\[\sqrt{41} = \sqrt{x^2+25}\]

Square both sides to eliminate the square roots:

\[41 = x^2 + 25\]

Subtract 25 from both sides to solve for \( x^2 \):

\[x^2 = 41 - 25 = 16\]

Now solve for \( x \) by taking the square root of both sides:

\[x = \pm\sqrt{16} = \pm4\]

Therefore, the values of \( x \) are \(\pm4\).