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 $M$.

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 $\frac{\sqrt{3}}{2}s$. Thus, the coordinates of $A$ are $\left( \frac{1+3}{2}, \frac{\sqrt{3}}{2}(3-1) \right) = (2, \sqrt{3})$.

Step 2: Find the equation of line AB

The line $AB$ passes through $A(2, \sqrt{3})$ and $B(1{,}0)$. The slope $m_{AB} = \frac{\sqrt{3}-0}{2-1} = \sqrt{3}$. Using the point-slope form, the equation of line $AB$ is $y - 0 = \sqrt{3}(x - 1)$, which simplifies to $y = \sqrt{3}x - \sqrt{3}$.

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 $M=(x_M, y_M)$ and $N=(x_N, y_N)$, then $x_M = \frac{0+x_N}{2}$ and $y_M = \frac{0+y_N}{2}$. This means $x_N = 2x_M$ and $y_N = 2y_M$. So, $N = (2x_M, 2y_M)$.

Step 4: Use the property of equilateral triangle and line ON

Since $\triangle ABC$ is equilateral, all its angles are $60^\circ$. The line $ON$ passes through the origin $O$. The line $AC$ passes through $A(2, \sqrt{3})$ and $C(3{,}0)$. The slope of $AC$ is $m_{AC} = \frac{\sqrt{3}-0}{2-3} = -\sqrt{3}$. The equation of line $AC$ is $y - 0 = -\sqrt{3}(x - 3)$, which simplifies to $y = -\sqrt{3}x + 3\sqrt{3}$. Consider $\triangle OAC$. The coordinates are $O(0{,}0)$, $A(2, \sqrt{3})$, $C(3{,}0)$. In $\triangle OBN$, $M$ 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 $M=(x_M, kx_M)$ and $N=(2x_M, 2kx_M)$. Since $M$ lies on $AB$, its coordinates satisfy $y_M = \sqrt{3}x_M - \sqrt{3}$. Since $N$ lies on $AC$, its coordinates satisfy $y_N = -\sqrt{3}x_N + 3\sqrt{3}$. Substitute $x_N = 2x_M$ and $y_N = 2y_M$ into the equation for line $AC$: $2y_M = -\sqrt{3}(2x_M) + 3\sqrt{3}$. This gives $2y_M = -2\sqrt{3}x_M + 3\sqrt{3}$. We have a system of two equations for $x_M$ and $y_M$: 1) $y_M = \sqrt{3}x_M - \sqrt{3}$ 2) $2y_M = -2\sqrt{3}x_M + 3\sqrt{3}$ Substitute (1) into (2): $2(\sqrt{3}x_M - \sqrt{3}) = -2\sqrt{3}x_M + 3\sqrt{3}$ $2\sqrt{3}x_M - 2\sqrt{3} = -2\sqrt{3}x_M + 3\sqrt{3}$ $4\sqrt{3}x_M = 5\sqrt{3}$ $x_M = \frac{5}{4}$ Now find $y_M$ using $y_M = \sqrt{3}x_M - \sqrt{3}$: $y_M = \sqrt{3}\left(\frac{5}{4}\right) - \sqrt{3} = \frac{5\sqrt{3}}{4} - \frac{4\sqrt{3}}{4} = \frac{\sqrt{3}}{4}$.

More from this chapter

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• Find the coordinates of the point which divides the line segment joining (4, -3) and (8, 5) in the ratio 3:1 internally.
• Find the relation between x and y if the point (x, y) is equidistant from (7, 1) and (3, 5).
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• Find the coordinates of the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4).
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