In Triangle ARP, centroid E is on median AX¯¯¯¯¯¯¯¯
. AE = y+6 and EX = 3y-2. Find AX
(1 point)
Responses
73
7 thirds
13
13
11
11
512
To find AX, we need to use the property of a centroid in a triangle. In a triangle, the centroid divides each median into two segments, with the segment closer to the vertex being twice as long as the segment closer to the midpoint of the opposite side.
Let's call the length of AX as a.
According to the property of a centroid,
AE = 2(EX)
Given that AE = y+6 and EX = 3y-2, we can set up the equation:
y+6 = 2(3y-2)
Simplifying the equation:
y+6 = 6y - 4
5y = 10
y = 2
Now, let's substitute the value of y back into the equation to find AX:
AE = y+6
AE = 2+6
AE = 8
Since AE is the sum of AX and EX, we can set up the equation:
8 = a + 3y - 2
Substituting the value of y:
8 = a + 3(2) - 2
8 = a + 6 - 2
8 = a + 4
Subtracting 4 from both sides:
a = 4
Therefore, AX is equal to 4.