Question

Find the distance between points P(3, -8) Q(7,4). Round to the nearest tenth if necessary

Answer 1

To calculate the distance between two points, we can use a formula that is a variation Pythagorean Theorem. Look:

$\mathsf{d=\sqrt{(x_Q-x_P)^2+(y_Q-y_P)^2}}$

"d" represents the distance and coordinates are expressed as follows: (x, y)

Let's go to the calculations.

$\mathsf{d=\sqrt{(x_Q-x_P)^2+(y_Q-y_P)^2}}\\\\ \mathsf{d=\sqrt{(7-3)^2+(4-(-8))^2}}\\\\ \mathsf{d=\sqrt{(4)^2+(4+8)^2}}\\\\ \mathsf{d=\sqrt{16+(12)^2}}\\\\ \mathsf{d=\sqrt{16+144}}\\\\ \mathsf{d=\sqrt{160}}\\\\ \mathsf{d=12,649110640673...}\\\\ \underline{\mathsf{d\approxeq12,6}}$

The answer is 12,6 u.c.

Answer 2

The distance between points P(3, -8) and Q(7, 4) on a coordinate plane is 12.7 units.

Given the following points:

• Points on the x-axis = (5, 1)

• Points on the y-axis = (5, 5)

To find the distance between the given points on a coordinate plane:

Mathematically, the distance between points on a coordinate plane is given by the formula:

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

Substituting the given points into the formula, we have;

$Distance = \sqrt{(7 - 3)^2 + (4 - [-8])^2}\\\\Distance = \sqrt{(4)^2 + (12)^2}\\\\Distance = \sqrt{16 + 144}\\\\Distance = \sqrt{160}$

Distance = 12.7 units.

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