Let `A(5, 12), B(-13 costheta, 13 sin theta) and C(13 sin theta, -13 cos theta)` are angular points of ABC where `theta in R`. The locus of orthocentre of DeltaABC is

Options

A. x-y=7
B. x-y+7=0
C. x+y-7=0
D. x+y+7=0

Explanation

To find the locus of the orthocenter of triangle ABC with vertices A(5, 12), B(-13 cos θ, 13 sin θ), and C(13 sin θ, -13 cos θ), we will follow these steps:

### Step 1: Find the coordinates of points B and C
Given:
- Point A: \( A(5, 12) \)
- Point B: \( B(-13 \cos \theta, 13 \sin \theta) \)
- Point C: \( C(13 \sin \theta, -13 \cos \theta) \)

### Step 2: Calculate the slope of line BC
The slope \( m_{BC} \) of line segment BC can be calculated using the formula:
\[ m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of B and C:
\[ m_{BC} = \frac{-13 \cos \theta - 13 \sin \theta}{13 \sin \theta - (-13 \cos \theta)} = \frac{-13 (\cos \theta + \sin \theta)}{13 (\sin \theta + \cos \theta)} = -1 \]

### Step 3: Find the slope of line AD
Since AD is perpendicular to BC, we have:
\[ m_{AD} = -\frac{1}{m_{BC}} = -\frac{1}{-1} = 1 \]

### Step 4: Write the equation of line AD
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (5, 12) \) and \( m = 1 \):
\[ y - 12 = 1(x - 5) \]
This simplifies to:
\[ y - 12 = x - 5 \implies x - y + 7 = 0 \]

### Step 5: Identify the locus of the orthocenter
The equation \( x - y + 7 = 0 \) represents a straight line. Since the orthocenter of triangle ABC lies on this line for all values of \( \theta \), the locus of the orthocenter is:
\[ x - y + 7 = 0 \]

### Conclusion
Thus, the locus of the orthocenter of triangle ABC is given by the line:
\[ x - y + 7 = 0 \]

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