he orthocentre of triangle formed by lines 4x−7y+10=0, x+y=54𝑥-7𝑦+10=0, 𝑥+𝑦=5 and 7x+4y=157𝑥+4𝑦=15 is

The orthocentre of a triangle is the point of intersection of the altitudes of the triangle. The altitudes of a triangle are the perpendiculars drawn from the vertices of the triangle to the opposite sides.

The given lines are:

1. 4x - 7y + 10 = 0
2. x + y = 5
3. 7x + 4y = 15

First, we need to find the points of intersection of these lines which will give us the vertices of the triangle.

Solving 1 and 2, we get the first vertex A. Solving 2 and 3, we get the second vertex B. Solving 1 and 3, we get the third vertex C.

Next, we need to find the equations of the altitudes. The altitude from a vertex will be perpendicular to the opposite side. So, we can find the slopes of the given lines and use the fact that the product of the slopes of two perpendicular lines is -1 to find the slopes of the altitudes.

Once we have the slopes of the altitudes, we can find their equations using the point-slope form of a line (y - y1 = m(x - x1)) where m is the slope and (x1, y1) is a point on the line (the vertex from which the altitude is drawn).

Finally, solving the equations of the altitudes will give us their points of intersection which is the orthocentre of the triangle.

Please note that this is a general method. The calculations can be quite complex depending on the coordinates of the vertices.