Question:
A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If \(\angle BAC=90^\circ\), and \(\mathrm{ar}(\Delta ABC)=5\sqrt{5}\) sq. units, then the abscissa of the vertex C is :
A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If \(\angle BAC=90^\circ\), and \(\mathrm{ar}(\Delta ABC)=5\sqrt{5}\) sq. units, then the abscissa of the vertex C is :
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AC is perpendicular to AB, so it has direction \((1,2)\); the area fixes the length AC.
Updated On: Nov 5, 2026
- \(1+2\sqrt{5}\)
- \(1+\sqrt{5}\)
- \(2+\sqrt{5}\)
- \(2\sqrt{5}-1\)
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The Correct Option is A
Solution and Explanation
\(\overrightarrow{AB}=(2,-1)\), \(|AB|=\sqrt{5}\). Since \(\angle BAC=90^\circ\), \(\overrightarrow{AC}\perp\overrightarrow{AB}\), so \(\overrightarrow{AC}=t(1,2)\).
Area \(=\frac12|AB||AC|=\frac12\sqrt{5}\cdot|t|\sqrt{5}=\frac{5|t|}{2}=5\sqrt{5}\Rightarrow |t|=2\sqrt{5}\).
\(C=(1+t,\ 2+2t)\). For C in the first quadrant, \(t=2\sqrt{5}\) (\(t=-2\sqrt{5}\) gives a negative abscissa).
Abscissa of C \(=1+2\sqrt{5}\).
Answer: (1) \(1+2\sqrt{5}\)
Area \(=\frac12|AB||AC|=\frac12\sqrt{5}\cdot|t|\sqrt{5}=\frac{5|t|}{2}=5\sqrt{5}\Rightarrow |t|=2\sqrt{5}\).
\(C=(1+t,\ 2+2t)\). For C in the first quadrant, \(t=2\sqrt{5}\) (\(t=-2\sqrt{5}\) gives a negative abscissa).
Abscissa of C \(=1+2\sqrt{5}\).
Answer: (1) \(1+2\sqrt{5}\)
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