If two vertices of a triangle are `(0,2)` and `(4,3)` and its orthocentre is `(0,0)` then the third vertex of the triangle lies in (a) `I^(st)` quadrant (b) `2^(nd)quadrant (c) `3^(rd)quadrant (d) `4^(th)quadrant

Options

A. Fourth
B. Second
C. Third
D. First

Answer

To find the coordinates of the third vertex of the triangle given the two vertices and the orthocenter, we will follow these steps:

### Step 1: Identify the given points
Let the vertices of the triangle be:
- A(0, 2)
- B(4, 3)
- C(H, K) (the third vertex we need to find)
- Orthocenter O(0, 0)

### Step 2: Find the slope of line AB
The slope of line AB can be calculated using the formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]
For points A(0, 2) and B(4, 3):
\[ \text{slope of AB} = \frac{3 - 2}{4 - 0} = \frac{1}{4} \]

### Step 3: Use the property of perpendicular lines
Since the orthocenter O(0, 0) is the intersection of the altitudes, line AB is perpendicular to line OC. Therefore, the product of their slopes must equal -1:
\[ \text{slope of AB} \cdot \text{slope of OC} = -1 \]
Let the slope of OC be \(\frac{K - 0}{H - 0} = \frac{K}{H}\). Thus:
\[ \frac{1}{4} \cdot \frac{K}{H} = -1 \]
This simplifies to:
\[ K = -4H \]

### Step 4: Find the slope of line AO
The slope of line AO (from A to O) is:
\[ \text{slope of AO} = \frac{0 - 2}{0 - 0} = \text{undefined} \quad \text{(vertical line)} \]
Since AO is vertical, line BC must be horizontal (slope = 0). Therefore:
\[ \text{slope of BC} = \frac{K - 3}{H - 4} = 0 \]
This implies:
\[ K - 3 = 0 \quad \Rightarrow \quad K = 3 \]

### Step 5: Substitute K back to find H
Now substitute \(K = 3\) into the equation \(K = -4H\):
\[ 3 = -4H \quad \Rightarrow \quad H = -\frac{3}{4} \]

### Step 6: Determine the coordinates of point C
The coordinates of the third vertex C are:
\[ C\left(-\frac{3}{4}, 3\right) \]

### Step 7: Identify the quadrant
The x-coordinate is negative and the y-coordinate is positive, which means point C lies in the second quadrant.

### Final Answer
The third vertex of the triangle lies in the **2nd quadrant**.

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