In the coordinate plane, triangle ABC is equilateral with B(1,0) and C(3,0). A line through the origin O meets AB and AC at M and N, respectively. If OM=MN, find the coordinates of M.
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Step-by-Step Solution
Step 1: Find the coordinates of A
Since triangle ABC is equilateral and B(1,0) and C(3,0) lie on the x -axis, the midpoint of BC is the x -coordinate of A. The length of the side BC is 3−1=2. The height of an equilateral triangle with side length s is 23s. Thus, the coordinates of A are (21+3,23(3−1))=(2,3).
Step 2: Find the equation of line AB
The line AB passes through A(2,3) and B(1,0). The slope mAB=2−13−0=3. Using the point-slope form, the equation of line AB is y−0=3(x−1), which simplifies to y=3x−3.
Step 3: Apply the condition OM = MN
Given that OM=MN, point M is the midpoint of the line segment ON. Since O is the origin (0,0), if M=(xM,yM) and N=(xN,yN), then xM=20+xN and yM=20+yN. This means xN=2xM and yN=2yM. So, N=(2xM,2yM).
Step 4: Use the property of equilateral triangle and line ON
Since △ABC is equilateral, all its angles are 60∘. The line ON passes through the origin O. The line AC passes through A(2,3) and C(3,0). The slope of AC is mAC=2−33−0=−3. The equation of line AC is y−0=−3(x−3), which simplifies to y=−3x+33.
Consider △OAC. The coordinates are O(0,0), A(2,3), C(3,0).
In △OBN, M is the midpoint of ON. B is (1,0). N is on AC. M is on AB. The line ON is a transversal. This problem can be solved using Menelaus' theorem or by using the property that M is the midpoint of ON. Let the line ON be y=kx. Then M=(xM,kxM) and N=(2xM,2kxM).
Since M lies on AB, its coordinates satisfy yM=3xM−3.
Since N lies on AC, its coordinates satisfy yN=−3xN+33.
Substitute xN=2xM and yN=2yM into the equation for line AC: 2yM=−3(2xM)+33.
This gives 2yM=−23xM+33.
We have a system of two equations for xM and yM:
1) yM=3xM−3
2) 2yM=−23xM+33
Substitute (1) into (2):
2(3xM−3)=−23xM+33 23xM−23=−23xM+33 43xM=53 xM=45
Now find yM using yM=3xM−3:
yM=3(45)−3=453−443=43.