Question:
If a \(\Delta ABC\) has vertices \(A(-1,7)\), \(B(-7,1)\) and \(C(5,-5)\), then its orthocentre has coordinates :
If a \(\Delta ABC\) has vertices \(A(-1,7)\), \(B(-7,1)\) and \(C(5,-5)\), then its orthocentre has coordinates :
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Intersect two altitudes: the altitude from \(C\) is perpendicular to \(AB\) and the altitude from \(A\) is perpendicular to \(BC\).
Updated On: Nov 5, 2026
- \((-3,3)\)
- \((3,-3)\)
- \(\left(\frac35,-\frac35\right)\)
- \(\left(-\frac35,\frac35\right)\)
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The Correct Option is A
Solution and Explanation
Slope of \(AB=\frac{1-7}{-7+1}=1\). Slope of \(BC=\frac{-5-1}{5+7}=-\frac12\).
Altitude from \(C\) (perpendicular to \(AB\), slope \(-1\)): \(y+5=-(x-5)\Rightarrow x+y=0\).
Altitude from \(A\) (perpendicular to \(BC\), slope \(2\)): \(y-7=2(x+1)\Rightarrow y=2x+9\).
Solving: \(x+2x+9=0\Rightarrow x=-3,\ y=3\).
Check with altitude from \(B\) (perpendicular to \(AC\), slope of \(AC=-2\), so slope \(\frac12\)): \(y-1=\frac12(x+7)\); at \((-3,3)\): \(2=2\). Verified.
Answer: (1) \((-3,3)\)
Altitude from \(C\) (perpendicular to \(AB\), slope \(-1\)): \(y+5=-(x-5)\Rightarrow x+y=0\).
Altitude from \(A\) (perpendicular to \(BC\), slope \(2\)): \(y-7=2(x+1)\Rightarrow y=2x+9\).
Solving: \(x+2x+9=0\Rightarrow x=-3,\ y=3\).
Check with altitude from \(B\) (perpendicular to \(AC\), slope of \(AC=-2\), so slope \(\frac12\)): \(y-1=\frac12(x+7)\); at \((-3,3)\): \(2=2\). Verified.
Answer: (1) \((-3,3)\)
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