Question

Which point on x-axis is equidistant from A(−4, 12) and B(−7, 9)?

Answer 1

Given:

A(−4, 12), B(−7, 9), and point P on x-axis with coordinates (x, 0).

Step 1: Since P is on the x-axis, its y-coordinate is 0.

So, P = (x, 0). 

Step 2: P is equidistant from A and B, so the distances PA and PB are equal.

`"Distance" = sqrt((x_2 − x_1)^2 + (y_2 − y_1)^2)`

So, \[PA = \sqrt{(x + 4)^2 + (0 - 12)^2} = \sqrt{(x + 4)^2 + 144}\]

\[PB = \sqrt{(x + 7)^2 + (0 - 9)^2} = \sqrt{(x + 7)^2 + 81}\]

Step 3: Equate PA and PB, \[\sqrt{(x + 4)^2 + 144} = \sqrt{(x + 7)^2 + 81}\]

(x + 4)2 + 144 = (x + 7)2 + 81   ...[Square both sides]

Step 4: Expand

(x² + 8x + 16) + 144 = (x² + 14x + 49) + 81

\[ x^2 + 8x + 160 = x^2 + 14x + 130\]

x² + 8x + 160 

= x² + 14x + 130

Step 5: 8x + 160 = 14x + 130 ... [Simplify]

160 − 130 = 14x − 8x

\[30 = 6x \implies x = \frac{30}{6} = 5\]

Therefore, the point on x-axis equidistant from A and B is at (5, 0).