Question

If two vertices of a triangle are \( (3,-2) \) and \( (-2,3) \) and its orthocentre is \( (-6,1) \). Then the difference between ordinate and abscissa of the third vertex of the triangle is

• A. 2
• B. 5
• C. \( -5 \)
• D. 7

Hint:

Use perpendicular slope property of altitudes: product of slopes = -1 to form equations.

Answer 1

The Correct Option is D

Step 1: Let the third vertex be \( (x,y) \).
Given vertices:
\[ A(3,-2),\quad B(-2,3),\quad C(x,y) \]
Orthocentre is:
\[ H(-6,1) \]

Step 2: Use property of orthocentre.
Orthocentre is the intersection of altitudes, so:
\[ AH \perp BC \quad \text{and} \quad BH \perp AC \]

Step 3: Find slope of \( AH \).
\[ m_{AH} = \frac{1-(-2)}{-6-3} = \frac{3}{-9} = -\frac{1}{3} \]
So, slope of \( BC \) is:
\[ m_{BC} = 3 \]

Step 4: Write slope of \( BC \).
\[ m_{BC} = \frac{y-3}{x+2} \]
So:
\[ \frac{y-3}{x+2} = 3 \]
\[ y - 3 = 3(x+2) \]
\[ y = 3x + 9 \quad \text{(1)} \]

Step 5: Find slope of \( BH \).
\[ m_{BH} = \frac{1-3}{-6+2} = \frac{-2}{-4} = \frac{1}{2} \]
So slope of \( AC \) is:
\[ m_{AC} = -2 \]

Step 6: Write slope of \( AC \).
\[ m_{AC} = \frac{y+2}{x-3} \]
\[ \frac{y+2}{x-3} = -2 \]
\[ y+2 = -2(x-3) \]
\[ y = -2x + 4 \quad \text{(2)} \]

Step 7: Solve equations (1) and (2).
\[ 3x + 9 = -2x + 4 \]
\[ 5x = -5 \Rightarrow x = -1 \]
\[ y = 3(-1) + 9 = 6 \]

Step 8: Find required difference.
\[ y - x = 6 - (-1) = 7 \]
Final Answer:
\[ \boxed{7} \]