NIMCET 2018 Question Paper with solutions

Calculation: Let's say that the lines are: $L_1 \rightarrow (1+p)x - py + p(1+p) = 0$ $L_2 \rightarrow (1+q)x - qy + q(1+q) = 0$ $L_3 \rightarrow y = 0$ Multiplying $L_1$ by $q$ and $L_2$ by $p$ and subtracting will give us: $x(q + pq - p - pq) + (pq + p^{2}q - pq - pq^{2}) = 0$ $\Rightarrow x(q-p) + pq(p-q) = 0$ $\Rightarrow x = pq.$ Substituting this in either $L_1$ or $L_2$, we will get: $(1+p)(pq) - py + p(1+p) = 0$ $\Rightarrow y = (1+p)(1+q)$ Therefore: $L_1$ and $L_2$ intersect at $A[pq, (1+p)(1+q)]$ . $L_1$ and $L_3$ intersect at $B(-p, 0)$ . $L_2$ and L3 intersect at C $(-q, 0)$ . Using the formula for line perpendicular to two points: Equation of altitude from A: $(-q+p)x + (0)y = (-q+p)(pq) + (0)(1+p)(1+q)$ $\Rightarrow x = pq$ Equation of altitude from B: $(pq+q)x + [(1+p)(1+q)-0]y = (pq+q)(-p) + [(1+p)(1+q)-0](0)$ $\Rightarrow q(1+p)x + (1+p)(1+q)y = (-pq)(1+p)$ $\Rightarrow qx + (1+q)y + pq = 0 \ldots (2)$ To find the ortho-center, we solve the equations (1) and (2) of the two altitudes: $q(pq) + (1+q)y + pq = 0$ $\Rightarrow (1+q)y + pq(1+q) = 0$ $\Rightarrow y = -pq$ $\therefore$ 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.