Question

Which of the following points has a distance of 3 from the point (-7, 4) A (-10, 4) B (-7, -1) C (4, 4) D (10, 4)

Answer 1

To find which point is 3 units away from the point \((-7, 4)\), we can use the distance formula, which is given by:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

where \((x_1, y_1)\) is the original point and \((x_2, y_2)\) is the point we are testing. We want the distance \(d\) to equal 3.

**Point A: (-10, 4)**
- \((x_1, y_1) = (-7, 4)\)
- \((x_2, y_2) = (-10, 4)\)

Calculating the distance:

\[
d = \sqrt{((-10) - (-7))^2 + (4 - 4)^2} = \sqrt{(-3)^2 + 0} = \sqrt{9} = 3
\]

**Point B: (-7, -1)**
- \((x_1, y_1) = (-7, 4)\)
- \((x_2, y_2) = (-7, -1)\)

Calculating the distance:

\[
d = \sqrt{((-7) - (-7))^2 + (-1 - 4)^2} = \sqrt{0 + (-5)^2} = \sqrt{25} = 5
\]

**Point C: (4, 4)**
- \((x_1, y_1) = (-7, 4)\)
- \((x_2, y_2) = (4, 4)\)

Calculating the distance:

\[
d = \sqrt{(4 - (-7))^2 + (4 - 4)^2} = \sqrt{(11)^2 + 0} = \sqrt{121} = 11
\]

**Point D: (10, 4)**
- \((x_1, y_1) = (-7, 4)\)
- \((x_2, y_2) = (10, 4)\)

Calculating the distance:

\[
d = \sqrt{(10 - (-7))^2 + (4 - 4)^2} = \sqrt{(17)^2 + 0} = \sqrt{289} = 17
\]

After checking all the points, we find that **Point A (-10, 4)** is the only point that has a distance of 3 from the point \((-7, 4)\).