Question

Relation between x and y such that the point (x, y) is equidistant from the points (-7, -1) and (3, 5) is :

• A. \(5x + y = 2\)
• B. \(5x + 3y = -4\)
• C. \(x - y = 2\)
• D. \(x - 2y = 2\)

Hint:

The locus of a point equidistant from two points is the perpendicular bisector of the segment joining them.
The midpoint of \(A(-7, -1)\) and \(B(3, 5)\) is:
\[ M = \left(\frac{-7+3}{2}, \frac{-1+5}{2}\right) = (-2, 2) \]
Since the midpoint must lie on this line, substitute \(x = -2, y = 2\) into the options:
For (B): \(5(-2) + 3(2) = -10 + 6 = -4\). This matches, confirming Option (B) is correct.

Answer 1

The Correct Option is B

Step 1: Think of this geometrically instead of algebraically.
A point equidistant from two fixed points $A$ and $B$ always lies on the perpendicular bisector of segment $AB$. So instead of expanding distance formulas, we can just find the equation of that perpendicular bisector directly for $A(-7,-1)$ and $B(3,5)$.
Step 2: Find the midpoint and slope of AB. \[ \text{Midpoint} = \left( \frac{-7+3}{2}, \frac{-1+5}{2} \right) = (-2, 2) \] \[ \text{Slope of } AB = \frac{5-(-1)}{3-(-7)} = \frac{6}{10} = \frac{3}{5} \]
Step 3: Take the negative reciprocal for the perpendicular slope and write the line through the midpoint. \[ \text{Slope of perpendicular} = -\frac{5}{3} \] \[ y - 2 = -\frac{5}{3}(x + 2) \]
Step 4: Clear the fraction and simplify. \[ 3(y-2) = -5(x+2) \] \[ 3y - 6 = -5x - 10 \] \[ 5x + 3y = -4 \] \[ \boxed{5x + 3y = -4} \]