Question:
Suppose \(ABOC\) is a rhombus in the first quadrant with \(O\) being the origin. If the vertices \(B\) and \(C\) of \(\triangle ABC\) lie respectively on
\[
y=\frac{4}{3}x
\]
and
\[
y=0,
\]
and the side \(BC\) passes through
\[
\left(\frac23,\frac23\right),
\]
then the midpoint of \(BC\) is
Suppose \(ABOC\) is a rhombus in the first quadrant with \(O\) being the origin. If the vertices \(B\) and \(C\) of \(\triangle ABC\) lie respectively on
\[
y=\frac{4}{3}x
\]
and
\[
y=0,
\]
and the side \(BC\) passes through
\[
\left(\frac23,\frac23\right),
\]
then the midpoint of \(BC\) is
Show Hint
In a rhombus, all sides are equal. Use the side-length condition first to express unknown coordinates, then apply the given geometric constraints to determine the parameters.
Updated On: Jun 26, 2026
- \(\left(\frac45,\frac25\right)\)
- \(\left(\frac23,\frac23\right)\)
- \(\left(\frac25,\frac45\right)\)
- \(\left(\frac13,\frac13\right)\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Represent the coordinates of \(B\) and \(C\).
Since \(B\) lies on \[ y=\frac43x, \] let \[ B=(3t,4t). \] Since \(C\) lies on the \(x\)-axis, \[ C=(a,0). \] Because \(ABOC\) is a rhombus with \(O=(0,0)\), all sides are equal. Hence \[ OB=OC. \] Now, \[ OB=\sqrt{(3t)^2+(4t)^2}=5t. \] Therefore, \[ OC=a=5t. \] Thus, \[ C=(5t,0). \]
Step 2: Use the fact that \(BC\) passes through \[ \left(\frac23,\frac23\right). \] The slope of \(BC\) is \[ m=\frac{0-4t}{5t-3t} =\frac{-4t}{2t} =-2. \] Hence the equation of \(BC\) is \[ y-4t=-2(x-3t). \] Substituting \[ \left(\frac23,\frac23\right), \] we get \[ \frac23-4t = -2\left(\frac23-3t\right). \] \[ \frac23-4t = -\frac43+6t. \] \[ 2=10t. \] \[ t=\frac15. \]
Step 3: Find the coordinates of \(B\) and \(C\).
\[ B= \left( \frac35,\frac45 \right), \] \[ C= \left( 1,0 \right). \]
Step 4: Find the midpoint of \(BC\).
The midpoint formula gives \[ M= \left( \frac{\frac35+1}{2}, \frac{\frac45+0}{2} \right). \] \[ M= \left( \frac{\frac85}{2}, \frac45\cdot\frac12 \right). \] \[ M= \left( \frac45, \frac25 \right). \]
Step 5: Final conclusion.
Therefore, the midpoint of \(BC\) is \[ \boxed{\left(\frac45,\frac25\right)} \] and the correct option is \[ \boxed{(1)}. \]
Since \(B\) lies on \[ y=\frac43x, \] let \[ B=(3t,4t). \] Since \(C\) lies on the \(x\)-axis, \[ C=(a,0). \] Because \(ABOC\) is a rhombus with \(O=(0,0)\), all sides are equal. Hence \[ OB=OC. \] Now, \[ OB=\sqrt{(3t)^2+(4t)^2}=5t. \] Therefore, \[ OC=a=5t. \] Thus, \[ C=(5t,0). \]
Step 2: Use the fact that \(BC\) passes through \[ \left(\frac23,\frac23\right). \] The slope of \(BC\) is \[ m=\frac{0-4t}{5t-3t} =\frac{-4t}{2t} =-2. \] Hence the equation of \(BC\) is \[ y-4t=-2(x-3t). \] Substituting \[ \left(\frac23,\frac23\right), \] we get \[ \frac23-4t = -2\left(\frac23-3t\right). \] \[ \frac23-4t = -\frac43+6t. \] \[ 2=10t. \] \[ t=\frac15. \]
Step 3: Find the coordinates of \(B\) and \(C\).
\[ B= \left( \frac35,\frac45 \right), \] \[ C= \left( 1,0 \right). \]
Step 4: Find the midpoint of \(BC\).
The midpoint formula gives \[ M= \left( \frac{\frac35+1}{2}, \frac{\frac45+0}{2} \right). \] \[ M= \left( \frac{\frac85}{2}, \frac45\cdot\frac12 \right). \] \[ M= \left( \frac45, \frac25 \right). \]
Step 5: Final conclusion.
Therefore, the midpoint of \(BC\) is \[ \boxed{\left(\frac45,\frac25\right)} \] and the correct option is \[ \boxed{(1)}. \]
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