If two vertical poles 20 m and 80 m high stand apart on a horizontal plane, then the height (in m) of the point of intersection of the lines joining the top of each pole to the foot of other is

Let the feet of the two poles be $$A$$ and $$B$$ on a horizontal line AB.
Height of pole at $$A$$ = $$20 \text{ m}$$, so its top is $$P(0,\,20)$$.
Height of pole at $$B$$ = $$80 \text{ m}$$, so its top is $$Q(d,\,80)$$, where $$d=AB$$ (the horizontal separation, whose value is not needed).

Choose a coordinate system with $$A(0,\,0)$$ and the positive $$x$$-axis along AB. Then:
• $$P(0,\,20)$$ is the top of the shorter pole.
• $$Q(d,\,80)$$ is the top of the taller pole.

The two required lines are:
1. $$PQ_1$$ joining $$P(0,\,20)$$ to the foot of the tall pole $$B(d,\,0)$$.
2. $$QP_1$$ joining $$Q(d,\,80)$$ to the foot of the short pole $$A(0,\,0)$$.

Equation of $$PB$$ (using two-point form):
$$\frac{y-20}{0-20}=\frac{x-0}{d-0} \;\Longrightarrow\; y = 20\!\left(1-\frac{x}{d}\right)$$.

Equation of $$QA$$ (straight line through the origin):
$$\frac{y-0}{80-0}=\frac{x-0}{d-0} \;\Longrightarrow\; y = 80\,\frac{x}{d}$$.

Let the intersection point of these two lines be $$R(x_0,\,y_0)$$. Equate the $$y$$-coordinates:

$$20\!\left(1-\frac{x_0}{d}\right)=80\,\frac{x_0}{d}$$
$$\Longrightarrow\; 20-20\,\frac{x_0}{d}=80\,\frac{x_0}{d}$$
$$\Longrightarrow\; 20 = 100\,\frac{x_0}{d}$$
$$\Longrightarrow\; \frac{x_0}{d}=0.2$$.

Substitute this ratio into $$y = 80\,\dfrac{x}{d}$$ to get the height of $$R$$:

$$y_0 = 80 \times 0.2 = 16$$.

Thus the two lines intersect at a height of $$16 \text{ m}$$ above the ground, independent of the horizontal separation of the poles.

Option A which is: 16