Calculation: Let's say that the lines are: L1→(1+p)x−py+p(1+p)=0 L2→(1+q)x−qy+q(1+q)=0 L3→y=0 Multiplying L1 by q and L2 by p and subtracting will give us: x(q+pq−p−pq)+(pq+p2q−pq−pq2)=0 ⇒x(q−p)+pq(p−q)=0 ⇒x=pq. Substituting this in either L1 or L2, we will get: (1+p)(pq)−py+p(1+p)=0 ⇒y=(1+p)(1+q) Therefore: L1 and L2 intersect at A[pq,(1+p)(1+q)] . L1 and L3 intersect at B(−p,0) . L2 and L3 intersect at C (−q,0) . Using the formula for line perpendicular to two points: Equation of altitude from A: (−q+p)x+(0)y=(−q+p)(pq)+(0)(1+p)(1+q) ⇒x=pq Equation of altitude from B: (pq+q)x+[(1+p)(1+q)−0]y=(pq+q)(−p)+[(1+p)(1+q)−0](0) ⇒q(1+p)x+(1+p)(1+q)y=(−pq)(1+p) ⇒qx+(1+q)y+pq=0...(2) To find the ortho-center, we solve the equations (1) and (2) of the two altitudes: q(pq)+(1+q)y+pq=0 ⇒(1+q)y+pq(1+q)=0 ⇒y=−pq ∴ The co-ordinates of the ortho-center are x=pq and y=−pq. The locus of the ortho-center will be x+y=0, which is a straight line.