Questions: Question 4 (Mandatory) (4 points) For the points P and Q, find the distance d(P, Q). P(6,-7), Q(2,-5) 6 12 2 sqrt(5) 12 sqrt(3)

Solution Steps

The coordinates of point $P$ are $(6, -7)$, and the coordinates of point $Q$ are $(2, -5)$.

The distance $d(P, Q)$ between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by: $$d(P, Q) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Substitute $x_1 = 6$, $y_1 = -7$, $x_2 = 2$, and $y_2 = -5$ into the formula: $$d(P, Q) = \sqrt{(2 - 6)^2 + (-5 - (-7))^2}$$

Calculate the differences: $$2 - 6 = -4 \quad \text{and} \quad -5 - (-7) = 2$$ Now, square these differences: $$(-4)^2 = 16 \quad \text{and} \quad 2^2 = 4$$

Add the squared differences: $$16 + 4 = 20$$ Take the square root of 20: $$\sqrt{20} = 2\sqrt{5}$$

Final Answer