Question:
In \(\Delta ABC\), if \(A = (1,2)\), and \(B\) and \(C\) lie on \(y = x + \alpha\) (where \(\alpha\) is variable), then the locus of the orthocenter of the triangle is
In \(\Delta ABC\), if \(A = (1,2)\), and \(B\) and \(C\) lie on \(y = x + \alpha\) (where \(\alpha\) is variable), then the locus of the orthocenter of the triangle is
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For triangles with two vertices moving on parallel lines, consider the altitude from the fixed vertex to find the orthocenter's locus.
Updated On: Jul 18, 2026
- \(x + y - 3 = 0\)
- \(x + y + 3 = 0\)
- \(y = x + 1\)
- \(y = x - 1\)
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The Correct Option is A
Solution and Explanation
Step 1: Set up coordinates.
Let \(A = (1,2)\), \(B = (x_1, x_1 + \alpha)\), \(C = (x_2, x_2 + \alpha)\).
Step 2: Slope of altitudes.
- Altitude from \(A\) is perpendicular to \(BC\).
- Slope of \(BC\) \(= \frac{(x_2 + \alpha) - (x_1 + \alpha)}{x_2 - x_1} = 1\)
- Therefore slope of altitude from \(A\) \(= -1\).
Step 3: Equation of altitude from A.
\[ y - 2 = -1(x - 1) \implies x + y - 3 = 0 \]
Step 4: Observation about orthocenter.
The orthocenter lies on this altitude because \(B\) and \(C\) are on \(y = x + \alpha\), and altitude from \(A\) is fixed.
Step 5: Locus of orthocenter.
Hence, the locus of the orthocenter is \[ \boxed{x + y - 3 = 0} \]
Let \(A = (1,2)\), \(B = (x_1, x_1 + \alpha)\), \(C = (x_2, x_2 + \alpha)\).
Step 2: Slope of altitudes.
- Altitude from \(A\) is perpendicular to \(BC\).
- Slope of \(BC\) \(= \frac{(x_2 + \alpha) - (x_1 + \alpha)}{x_2 - x_1} = 1\)
- Therefore slope of altitude from \(A\) \(= -1\).
Step 3: Equation of altitude from A.
\[ y - 2 = -1(x - 1) \implies x + y - 3 = 0 \]
Step 4: Observation about orthocenter.
The orthocenter lies on this altitude because \(B\) and \(C\) are on \(y = x + \alpha\), and altitude from \(A\) is fixed.
Step 5: Locus of orthocenter.
Hence, the locus of the orthocenter is \[ \boxed{x + y - 3 = 0} \]
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