Triangle L N P has centroid S. Lines are drawn from each point to the midpoint of the opposite side to form line segments N R, L P, and Q M. The length of line segment N S is 7 x minus 3 and the length of line segment S R is 5 x minus 3.

Question

What is the length of segment NS?

1 unit
2 units
4 units
6 units

Answer 1

To find the length of segment NS, we must recognize the properties of a triangle's centroid. The centroid (denoted as S) divides each median in a specific ratio; it divides the median into two segments where one segment (the part from the vertex to the centroid) is twice the length of the other segment (the part from the centroid to the midpoint of the opposite side).

Given:

• Length of segment NS = \( 7x - 3 \)
• Length of segment SR = \( 5x - 3 \)

Since S is the centroid, we know that: \[ NS = 2 \cdot SR \]

We can set up the equation: \[ 7x - 3 = 2(5x - 3) \]

Now, expanding the right side: \[ 7x - 3 = 10x - 6 \]

Now, we can rearrange this equation to solve for x:

1. Subtract \( 7x \) from both sides: \[ -3 = 3x - 6 \]
2. Add \( 6 \) to both sides: \[ 3 = 3x \]
3. Divide by \( 3 \): \[ x = 1 \]

Now we can substitute \( x \) back into the expression for NS to find its length: \[ NS = 7x - 3 = 7(1) - 3 = 7 - 3 = 4 \]

Thus, the length of segment NS is: \[ \boxed{4 \text{ units}} \]