Question:
For a triangle formed by \((0,0)\), \((4,0)\), and \((3,4)\), the orthocenter is
For a triangle formed by \((0,0)\), \((4,0)\), and \((3,4)\), the orthocenter is
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The orthocenter is the intersection of altitudes. Use perpendicular slopes from vertices to opposite sides and solve for intersection.
Updated On: Jul 18, 2026
- \(\left(3, \frac{3}{4}\right)\)
- \(\left(3, \frac{5}{4}\right)\)
- \((3,12)\)
- \((3,9)\)
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The Correct Option is A
Solution and Explanation
Step 1: Coordinates of vertices.
Let \(A = (0,0)\), \(B = (4,0)\), \(C = (3,4)\).
Step 2: Find slopes of sides.
\(\text{Slope of BC} = \frac{4-0}{3-4} = -4\)
\(\text{Slope of AC} = \frac{4-0}{3-0} = \frac{4}{3}\)
Step 3: Slopes of altitudes.
Altitude from A perpendicular to BC: slope \(= \frac{1}{4}\)
Equation: \(y - 0 = \frac{1}{4}(x - 0) \implies y = \frac{x}{4}\)
Step 4: Altitude from B perpendicular to AC.
Slope \(= -\frac{3}{4}\)
Equation: \(y - 0 = -\frac{3}{4}(x - 4) \implies y = -\frac{3}{4}x + 3\)
Step 5: Solve for intersection (orthocenter).
\(\frac{x}{4} = -\frac{3}{4}x + 3 \implies x = 3, y = \frac{3}{4}\)
Step 6: Final conclusion.
Hence, the orthocenter is \[ \boxed{\left(3, \frac{3}{4}\right)} \]
Let \(A = (0,0)\), \(B = (4,0)\), \(C = (3,4)\).
Step 2: Find slopes of sides.
\(\text{Slope of BC} = \frac{4-0}{3-4} = -4\)
\(\text{Slope of AC} = \frac{4-0}{3-0} = \frac{4}{3}\)
Step 3: Slopes of altitudes.
Altitude from A perpendicular to BC: slope \(= \frac{1}{4}\)
Equation: \(y - 0 = \frac{1}{4}(x - 0) \implies y = \frac{x}{4}\)
Step 4: Altitude from B perpendicular to AC.
Slope \(= -\frac{3}{4}\)
Equation: \(y - 0 = -\frac{3}{4}(x - 4) \implies y = -\frac{3}{4}x + 3\)
Step 5: Solve for intersection (orthocenter).
\(\frac{x}{4} = -\frac{3}{4}x + 3 \implies x = 3, y = \frac{3}{4}\)
Step 6: Final conclusion.
Hence, the orthocenter is \[ \boxed{\left(3, \frac{3}{4}\right)} \]
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