Question

If the point $P(x,y)$ is equidistant from the points $(3,7)$ and $(5,-1)$, then $4y-x=$

• A. $-8$
• B. $16$
• C. $8$
• D. $-16$

Hint:

Whenever a point is equidistant from two given points, immediately equate the squared distances instead of using square roots. This simplifies the algebra considerably: \[ (x-3)^2+(y-7)^2=(x-5)^2+(y+1)^2. \] After simplification, \[ x-4y=-8 \] which directly gives \[ 4y-x=8. \]

Answer 1

The Correct Option is C

Concept: When a point is equidistant from two fixed points, the distance from that point to the first point is equal to the distance from that point to the second point. The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: \[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. \] To avoid square roots, we usually equate the squares of the distances.

Step 1: Form the distance equation. The given point is \[ P(x,y). \] The two fixed points are \[ A(3,7) \] and \[ B(5,-1). \] Since $P$ is equidistant from $A$ and $B$, \[ PA=PB. \] Squaring both sides, \[ PA^2=PB^2. \] Substituting the coordinates, \[ (x-3)^2+(y-7)^2=(x-5)^2+(y+1)^2. \]

Step 2: Expand both sides completely. Expanding the left-hand side: \[ (x-3)^2+(y-7)^2 \] \[ =x^2-6x+9+y^2-14y+49 \] \[ =x^2+y^2-6x-14y+58. \] Expanding the right-hand side: \[ (x-5)^2+(y+1)^2 \] \[ =x^2-10x+25+y^2+2y+1 \] \[ =x^2+y^2-10x+2y+26. \] Therefore, \[ x^2+y^2-6x-14y+58 = x^2+y^2-10x+2y+26. \]

Step 3: Simplify the equation. Canceling $x^2$ and $y^2$ from both sides, \[ -6x-14y+58 = -10x+2y+26. \] Bringing all variable terms to one side, \[ -6x+10x-14y-2y = 26-58. \] \[ 4x-16y=-32. \] Dividing throughout by $4$, \[ x-4y=-8. \]

Step 4: Find the required expression. We need to evaluate \[ 4y-x. \] From \[ x-4y=-8, \] multiplying by $-1$ gives \[ 4y-x=8. \] Therefore, \[ \boxed{4y-x=8}. \] Hence, the correct answer is \[ \boxed{(C)\ 8}. \]