Equation of an Altitude

Step 1: Find the Vertices of the Triangle
By solving the line equations in pairs, we find the three intersection points (vertices):
- Vertex A (Intersection of $$x+y=7$$ and $$3x+4y=25$$): $$(3, 4)$$
- Vertex B (Intersection of $$3x+4y=25$$ and $$x+2y=10$$): $$(5, 2.5)$$
- Vertex C (Intersection of $$x+y=7$$ and $$x+2y=10$$): $$(4, 3)$$

Step 2: Find the Equations of Two Altitudes
The orthocentre is the intersection point of the altitudes.

- Altitude from Vertex A $$(3,4)$$ to side BC ($$x+2y=10$$):
- The slope of side BC is $$-\frac{1}{2}$$.
- The slope of the perpendicular altitude is $$2$$.
- Equation: $$y - 4 = 2(x - 3)$$ which simplifies to $$2x - y = 2$$

- Altitude from Vertex C $$(4,3)$$ to side AB ($$3x+4y=25$$):
- The slope of side AB is $$-\frac{3}{4}$$.
- The slope of the perpendicular altitude is $$\frac{4}{3}$$.
- Equation: $$y - 3 = \frac{4}{3}(x - 4)$$ which simplifies to $$4x - 3y = 7$$

Step 3: Find the Intersection of the Altitudes
Solve the system of equations from the two altitudes:
1. $$y = 2x - 2$$
2. $$4x - 3y = 7$$

Substitute equation (1) into equation (2):
$$4x - 3(2x - 2) = 7$$
$$4x - 6x + 6 = 7$$
$$-2x = 1$$
$$x = -0.5$$

Substitute $$x = -0.5$$ back into equation (1):
$$y = 2(-0.5) - 2$$
$$y = -1 - 2$$
$$y = -3$$

Therefore, the coordinates of the orthocentre are $$(-0.5, -3)$$.