NIMCET 2018 — Mathematics PYQ

Concept:
- Ortho-center of a triangle is the point of concurrence of its altitudes (heights).
- The equation of the line passing through a point (a, b), perpendicular to the line passing through (x₁, y₁) and (x₂, y₂), is:
$$(x_2-x_1)x+(y_2-y_1)y=(x_2-x_1)a+(y_2-y_1)b$$

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 + p^2q - pq - pq^2) = 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:
$$(-q+p)x+(0)y=(-q+p)(pq)+(0)(1+p)(1+q)$$
$$<br>\Rightarrow x=pq$$
$$\dots (1)$$
Equation of altitude from B:
$$<br>(pq+q)x+[(1+p)(1+q)-0]y=(pq+q)(-p)+[(1+p)(1+q)-0](0)$$
$$\Rightarrow qx+(1+q)y=(-pq)(1+p)$$
$$<br>\Rightarrow qx+(1+q)y+pq=0$$
$$\dots (2)$$
To find the ortho-center, we solve the equations (1) and (2) of the two altitudes:
$$<br>q(pq)+(1+q)y+pq=0$$
$$\Rightarrow (1+q)y+pq(1+q)=0$$
$$<br>\Rightarrow y=-pq$$
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.