Question

If (2,-6),(5,2) and (-2,2) constitute the vertices of a triangle then the the line joining the origin and its orthocentre is

• A. x+4y=0
• B. x-4y=0
• C. 4x-y=0
• D. 4x+y=0
• E. x-y=0

Answer 1

The Correct Option is B

Let the vertices be A(2, -6), B(5, 2), and C(-2, 2).

1. Slope of AB: \[ m_{AB} = \frac{2 - (-6)}{5 - 2} = \frac{8}{3} \]

2. Slope of BC: \[ m_{BC} = \frac{2 - 2}{-2 - 5} = \frac{0}{-7} = 0 \] (BC is horizontal)

3. Slope of AC: \[ m_{AC} = \frac{2 - (-6)}{-2 - 2} = \frac{8}{-4} = -2 \]

4. Altitude from C to AB: The altitude from C is perpendicular to AB. The slope of this altitude is the negative reciprocal of \(m_{AB}\): \[ m_{altC} = -\frac{3}{8} \] The equation of the altitude from C is \( y - 2 = -\frac{3}{8}(x + 2) \).

5. Altitude from A to BC: Since BC is horizontal, the altitude from A is vertical. Its equation is simply \( x = 2 \).

6. Find the orthocenter: The orthocenter is the intersection of the altitudes. Substitute \(x = 2\) into the equation of the altitude from C:

\[ y - 2 = -\frac{3}{8}(2 + 2) \] \[ y - 2 = -\frac{3}{8}(4) \] \[ y - 2 = -\frac{3}{2} \] \[ y = 2 - \frac{3}{2} = \frac{1}{2} \] So, the orthocenter is H(2, 1/2).

7. Line joining origin and orthocenter: The origin is O(0, 0). The line joining O and H has slope: \[ m_{OH} = \frac{1/2 - 0}{2 - 0} = \frac{1}{4} \] Using the point-slope form with the origin, the equation of the line is \[ y = \frac{1}{4}x \] or \[ x - 4y = 0 \]

Therefore, the answer is (B).

Answer 2

Step 1: Understand the problem and given vertices:

• The vertices of the triangle are \( A(2, -6) \), \( B(5, 2) \), and \( C(-2, 2) \).
• We are tasked with finding the equation of the line joining the origin \((0, 0)\) and the orthocenter of the triangle.

Step 2: Recall properties of the orthocenter:

• The orthocenter is the point of intersection of the altitudes of a triangle.
• An altitude is a perpendicular line drawn from a vertex to the opposite side (or its extension).

Step 3: Find slopes of sides of the triangle:

• Slope of \( AB \):
  • \( m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-6)}{5 - 2} = \frac{8}{3} \).
• Slope of \( BC \):
  • \( m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 2}{-2 - 5} = 0 \).
• Slope of \( CA \):
  • \( m_{CA} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - 2}{2 - (-2)} = \frac{-8}{4} = -2 \).

Step 4: Find equations of altitudes:

• Altitude from \( A \) to \( BC \):
  • Since \( BC \) is horizontal (\( m_{BC} = 0 \)), the altitude from \( A \) will be vertical. Its equation is:
  • \( x = 2 \).
• Altitude from \( B \) to \( CA \):
  • The slope of \( CA \) is \( m_{CA} = -2 \), so the slope of the perpendicular line is \( \frac{1}{2} \).
  • Using the point-slope form of a line through \( B(5, 2) \):
  • \( y - 2 = \frac{1}{2}(x - 5) \).
  • Simplify:
  • \( y - 2 = \frac{1}{2}x - \frac{5}{2} \).
  • \( 2y - 4 = x - 5 \).
  • \( x - 2y = 1 \).
• Altitude from \( C \) to \( AB \):
  • The slope of \( AB \) is \( m_{AB} = \frac{8}{3} \), so the slope of the perpendicular line is \( -\frac{3}{8} \).
  • Using the point-slope form of a line through \( C(-2, 2) \):
  • \( y - 2 = -\frac{3}{8}(x + 2) \).
  • Simplify:
  • \( y - 2 = -\frac{3}{8}x - \frac{6}{8} \).
  • \( 8y - 16 = -3x - 6 \).
  • \( 3x + 8y = 10 \).

Step 5: Solve for the orthocenter:

• The orthocenter is the intersection of the altitudes \( x = 2 \), \( x - 2y = 1 \), and \( 3x + 8y = 10 \).
• Substitute \( x = 2 \) into \( x - 2y = 1 \):
  • \( 2 - 2y = 1 \).
  • \( -2y = -1 \).
  • \( y = \frac{1}{2} \).
• Thus, the orthocenter is \( (2, \frac{1}{2}) \).

Step 6: Find the equation of the line joining the origin and the orthocenter:

• The slope of the line joining \( (0, 0) \) and \( (2, \frac{1}{2}) \) is:
  • \( m = \frac{\frac{1}{2} - 0}{2 - 0} = \frac{1}{4} \).
• Using the point-slope form of a line through \( (0, 0) \):
  • \( y = \frac{1}{4}x \).
  • Rewriting in standard form:
  • \( x - 4y = 0 \).

Final Answer:

The correct option is (B) \( x - 4y = 0 \).