Question:
A triangle has a vertex at \((1,2)\) and the midpoints of two sides through it are \((-1,1)\) and \((2,3)\). Find its area.
A triangle has a vertex at \((1,2)\) and the midpoints of two sides through it are \((-1,1)\) and \((2,3)\). Find its area.
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If a midpoint and one endpoint are known, the other endpoint is obtained using
\[
(x,y)=(2x_m-x_1,\;2y_m-y_1).
\]
Updated On: Jun 11, 2026
- \(1\)
- \(2\)
- \(3\)
- \(4\)
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The Correct Option is B
Solution and Explanation
Step 1: Determine the remaining vertices.
Let the given vertex be \[ A=(1,2). \] If midpoint of \(AB\) is \((-1,1)\), then \[ B=(2(-1)-1,\;2(1)-2) = (-3,0). \] If midpoint of \(AC\) is \((2,3)\), then \[ C=(2(2)-1,\;2(3)-2) = (3,4). \]
Step 2: Apply the coordinate area formula.
\[ \text{Area} = \frac12 \left| x_1(y_2-y_3) +x_2(y_3-y_1) +x_3(y_1-y_2) \right|. \] Substituting, \[ = \frac12 \left| 1(0-4) +(-3)(4-2) +3(2-0) \right| \] \[ = \frac12|-4-6+6| = \frac12(4) = 2. \] Hence \[ \boxed{2}. \]
Let the given vertex be \[ A=(1,2). \] If midpoint of \(AB\) is \((-1,1)\), then \[ B=(2(-1)-1,\;2(1)-2) = (-3,0). \] If midpoint of \(AC\) is \((2,3)\), then \[ C=(2(2)-1,\;2(3)-2) = (3,4). \]
Step 2: Apply the coordinate area formula.
\[ \text{Area} = \frac12 \left| x_1(y_2-y_3) +x_2(y_3-y_1) +x_3(y_1-y_2) \right|. \] Substituting, \[ = \frac12 \left| 1(0-4) +(-3)(4-2) +3(2-0) \right| \] \[ = \frac12|-4-6+6| = \frac12(4) = 2. \] Hence \[ \boxed{2}. \]
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