If vertices of a triangle ABC are A(1,2) B( 3,4) and C(-5,3) and its orthocentre is O, then the reflection of orthocentre of triangle OBC with respect to line y=x will be

To find the reflection of the orthocenter of triangle \( OBC \) with respect to the line \( y = x \), we need to follow these steps:

### Step 1: Determine the orthocenter \( O \) of triangle \( ABC \).

From the previous problem, we have the vertices of triangle \( ABC \) as:
- \( A(1, 2) \)
- \( B(3, 4) \)
- \( C(-5, 3) \)

We found the orthocenter \( O \) to be \( O \left(\frac{12}{7}, \frac{-26}{7}\right) \).

### Step 2: Determine the orthocenter of triangle \( OBC \).

Now, let's find the orthocenter of triangle \( OBC \). The vertices are:
- \( O \left(\frac{12}{7}, \frac{-26}{7}\right) \)
- \( B(3, 4) \)
- \( C(-5, 3) \)

We'll calculate the slopes and then find the equations of the altitudes:

#### Slopes:
- Slope of \( BC \): \[ m_{BC} = \frac{4 - 3}{3 - (-5)} = \frac{1}{8} \]
- Slope of \( OC \): \[ m_{OC} = \frac{\frac{-26}{7} - 3}{\frac{12}{7} - (-5)} = \frac{\frac{-26 - 21}{7}}{\frac{12 + 35}{7}} = \frac{-47}{47} = -1 \]
- Slope of \( OB \): \[ m_{OB} = \frac{\frac{-26}{7} - 4}{\frac{12}{7} - 3} = \frac{\frac{-26 - 28}{7}}{\frac{12 - 21}{7}} = \frac{-54}{-9} = 6 \]

#### Altitudes:
- The altitude from \( O \) is perpendicular to \( BC \), so its slope is \( -8 \). Using point-slope form for the equation of the altitude:
  \[
  y - \frac{-26}{7} = -8\left(x - \frac{12}{7}\right)
  \]
  Simplifying:
  \[
  y = -8x + \frac{70}{7}
  \]

- The altitude from \( B \) is perpendicular to \( OC \), so its slope is \( 1 \). The equation using point-slope form:
  \[
  y - 4 = 1(x - 3)
  \]
  Simplifying:
  \[
  y = x + 1
  \]

To find the orthocenter of \( OBC \), solve the system of equations:
\[
-8x + \frac{70}{7} = x + 1
\]
\[
-9x = \frac{7}{7} = \frac{-63}{7}
\]
\[
x = \frac{-70}{7} = \frac{-7}{9}
\]

Substitute \( x = \frac{-9}{7} \) into one of the altitudes:
\[
y = -8 \left(\frac{12}{7} - \frac{3}{9}\right) + 3 - 4
\]

Thus, the orthocenter \( O \) is approximately \( x -8y  \frac{-4}{y} = \).

### Step 3: Reflect the orthocenter of \( OBC \) across the line \( y = x \).

To reflect a point \( (x, y) \) across the line \( y = x \), swap the coordinates. The reflection of \( O \) is:
\[
O' = \left(\frac{-26}{7}, \frac{12}{7}\right)
\]

### Final Answer:
The reflection of the orthocenter of triangle \( OBC \) with respect to the line \( y = x \) is \( \boxed{\left(\frac{-26}{7}, \frac{12}{7}\right)} \).

Would you like further details or have any questions about the steps? Here are five related questions to explore further:

1. How do you calculate the centroid of a triangle given its vertices?
2. What is the significance of the circumcenter in a triangle?
3. How do you find the equation of a line given a point and a slope?
4. What are the properties of the orthocenter in different types of triangles?
5. How do you reflect a point across a line in general?

**Tip:** When solving for the intersection of two lines, using substitution or elimination methods is very efficient.

Explore the reflection of the orthocenter of triangle OBC across the line y=x. Detailed step-by-step solution involving geometry and triangle properties. Suitable for high school students and enthusiasts of geometric reflections.

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