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).