Question

Show that P(1,4) is equidistant from A(-5,-3) and B(-1,-5).

a) P is equidistant
b) P is not equidistant
c) Insufficient information
d) None of the above

Answer 1

By calculating the distances PA and PB using the distance formula, it's shown that point P(1,4) is equidistant from points A(-5,-3) and B(-1,-5), both being √[85], hence P is equidistant from A and B.

Step-by-step explanation:

To determine if point P(1,4) is equidistant from points A(-5,-3) and B(-1,-5), we calculate the distances PA and PB using the distance formula d = √[(x2-x1)² + (y2-y1)²].

For distance PA:

1. Substitute the coordinates into the distance formula: d = √[(1-(-5))² + (4-(-3))²].
2. Simplify the expression: d = √[(1+5)² + (4+3)²] = √[6² + 7²].
3. Calculate the result: d = √[36 + 49] = √[85].

For distance PB:

1. Substitute the coordinates into the distance formula: d = √[(1-(-1))² + (4-(-5))²].
2. Simplify the expression: d = √[(1+1)² + (4+5)²] = √[2² + 9²].
3. Calculate the result: d = √[4 + 81] = √[85].

Since PA and PB are both √[85], point P is indeed equidistant from points A and B. Therefore, the answer is (a) P is equidistant.