Let the equations of perpendicular bisectors of sides `AC and of Delta ABCbe x+y=3 and x-y=1` respectively Then vertex A is is (0,0)
The orthocentre of `DeltaABC` is

Options

A. (A)`(1,1)`
B. (B)`(2,1)`
C. (C)`(0,0)`
D. (D)`(0,1)`

Answer

To find the orthocenter of triangle ABC given the equations of the perpendicular bisectors of sides AC and AB, we can follow these steps:

### Step 1: Identify the equations of the perpendicular bisectors
The equations given are:
1. Perpendicular bisector of AC: \( x + y = 3 \)
2. Perpendicular bisector of AB: \( x - y = 1 \)

### Step 2: Find the intersection point of the perpendicular bisectors
The orthocenter of the triangle is the point where the altitudes intersect. Since we have the equations of the perpendicular bisectors, we can find their intersection point by solving the two equations simultaneously.

1. From the first equation \( x + y = 3 \), we can express \( y \) in terms of \( x \):
\[ y = 3 - x \]

2. Substitute \( y \) in the second equation \( x - y = 1 \):
\[ x - (3 - x) = 1 \]
\[ x - 3 + x = 1 \]
\[ 2x - 3 = 1 \]
\[ 2x = 4 \]
\[ x = 2 \]

3. Now substitute \( x = 2 \) back into the equation for \( y \):
\[ y = 3 - 2 = 1 \]

Thus, the intersection point (orthocenter) is \( (2, 1) \).

### Step 3: Conclusion
The orthocenter of triangle ABC is at the point \( (2, 1) \).

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