Question

Determine the value(s) of n.

P(3n, n-7), Q(4n, n+5), PQ=13

A. n = 3
B. n = 5
C. n = 7
D. n = 9

Answer 1

The correct value of n for which the distance between the points P(3n, n-7) and Q(4n, n+5) is 13 units is n = 5.

Step-by-step explanation:

The student is asked to determine the value(s) of n for which the distance between the points P(3n, n-7) and Q(4n, n+5) is 13 units. To find this, we apply the distance formula between the two points:

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

Plugging in the coordinates of P and Q, we get:

13 = \(\sqrt{(4n - 3n)^2 + ((n + 5) - (n - 7))^2}\)

Simplifying the equation:

13 = \(\sqrt{n^2 + 12^2}\)

13 = \(\sqrt{n^2 + 144}\)

Squaring both sides gives:

169 = n^2 + 144

n^2 = 169 - 144

n^2 = 25

n = \(\pm\sqrt{25}\)

n = \(\pm5\)

However, since n represents a coordinate, we consider only the positive value, thus n = 5.