Question

5
4. A point (x,y) is equidistant from the points C (6, 8) and F (6, 1). Then
(a)2x -7y +36 =0 (b)14y = 63 (c)x-y = 5 (d) x + y=5​

Answer 1

Answer:

(b) $14y = 63$

Step-by-step explanation:

Let the points be A (x, y), C (6, 8) and F (6, 1).

Using Distance Formula:

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

$AC = \sqrt{(6-x)^2 + (8-y)^2}$

$AF = \sqrt{(6-x)^2 + (1-y)^2}$

Given that (x, y) is equidistant from (6, 8) and (6, 1)  i.e.  AC = AF.

∴ $\sqrt{(6-x)^2 + (8-y)^2} = \sqrt{(6-x)^2 + (1-y)^2}$

Squaring both sides:

$(\sqrt{(6-x)^2 + (8-y)^2})^2 = (\sqrt{(6-x)^2 + (1-y)^2})^2$

$(6-x)^2 + (8-y)^2 = (6-x)^2 + (1-y)^2$

Cancel $(6-x)^2$ on both sides:

$(8-y)^2 = (1-y)^2$

Expand using $(a+b)^2 = a^2 + 2ab + b^2$:

$(64 - 16y + y^2) = (1 - 2y + y^2)$

Cancel $y^2$ on both sides:

$64 - 16 y = 1 - 2y$

Transpose all terms to the LHS:

$63 - 14y = 0$

Bring 14y to the RHS:

$63 = 14y$

∴  $14y = 63$ is the equation of all points (x, y) equidistant from (6, 8) and (6, 1)

Shortcut:

"Locus of a point equidistant from two points is the perpendicular bisector of the line joining those two points."

point of intersection of $\perp$ bisector and CF = midpoint of CF

$(x_1, y_1)$ = $(\frac{6+6}{2} , \frac{8+1}{2} ) = (6, \frac{9}{2})$

CF is a straight line along x = 6 => slope = tan 90°

Slope of $\perp$ bisector = m = tan 0° = 0

∴ Eqn. of $\perp$ bisector = $(y-y_1) = m(x-x_1)$ = (y - $\frac{9}{2}$) = 0 (x - 6)  =>  $y = \frac{9}{2} = \frac{63}{14}$  

Any corrections or suggestions for improvement are welcome :)