### Q.3. (i) Prove that the elasticity of a downward sloping straight-line demand curve varies from infinity \(\left ( \infty \right )\) at the price- axis to zero at the quantity-axis.

### (ii) Comparing two straight line demand curves of the same slope, the one farther from the origin is less elastic at each price than the one closer to the origin. Explain.

### (iii) Prove that two intersecting straight line demand curves have different price elasticities at the point of intersection.

### (iv) Prove that in the case of two straight line demand curves, with the same point of origin on the price-axis, at any given price elasticity is the same inspite of different slopes.

Ans. (i) The slope of a straight line having a negative slope is the same throughout. This means \(\frac{{\Delta p}}{{\Delta q}}\) is the same throughout. This further means that its inverse \(\frac{{\Delta q}}{{\Delta p}}\) is also the same or constant throughout. Therefore, changes in \({e_d}\) can be found out by comparing \(\frac{p}{q}\). The point where the *dd* line cuts Y-axis, \(\frac{p}{q}\) is \(\infty\) because quantity is zero. As we move down the line, p falls and q rises steadily thus \(\frac{p}{q}\) is falling steadily so that \({e_d}\) is also falling. On the x-axis, the price is zero. So the ratio \(\frac{p}{q}\) is zero. Thus \({e_d}\) = 0.

(ii) The two straight-line demand curves of the same slope are parallel lines. Now we take price OP and compare the elasticities of the two curves at this price. Since the curves are parallel, the ratio \(\frac{{\Delta q}}{{\Delta p}}\) is the same on both the curves. Since we are comparing the elasticities on the same price, so *p* is also the same. Now one thing, which varies, is *q*. On the curve, farther from the origin, quantity is larger i.e., \(Qq_{2}> Oq_{1}\) and hence \(\frac{p}{q}\) is smaller and thus\(e\) is smaller.

Thus, \(e\) shall be smaller in those cases where demand curves of the same slope are farther from the origin.

(iii) AB and CD intersect at point E. On this point at lines AB and CD, \({e_d}\) is \[\frac{\Delta Q}{\Delta P}\times \frac{P}{Q}=\frac{QQ_{I}}{PP_{I}}\times \frac{OP}{OQ}\] and \[\frac{QQ_{2}}{PP_{1}}\times \frac{OP}{OQ}\]

\(\frac{OP}{OQ}\) is common in both the case.

\(\therefore\) \({e_d}\) can be compared by comparing \(\frac{QQ_{1}}{PP_{1}}\) and \(\frac{QQ_{2}}{PP_{1}}\).

But \(\frac{QQ_{1}}{PP_{1}}\neq \frac{QQ_{2}}{PP_{1}}\)

This means \({e_d}\) at the point of intersection is not the same on both the lines having different slopes.

\((iv)\,\,ed = \frac{{\Delta Q}}{{\Delta P}} \times \frac{P}{Q}\)

Hence, \(\frac{\Delta Q}{\Delta P}\) is the inverse of the slope and in its place, we can use the measure of the slope. Now, \({e_d}\) in both the cases would be as follows:

\(Case\, – I:\frac{{OM}}{{O{P_1}}} \times \frac{{OP}}{{OQ}}\,\,or\,\frac{{OP}}{{O{P_1}}} \times \frac{{OM}}{{O{Q_1}}}\)

\(Case\, – II:\frac{{ON}}{{O{P_1}}} \times \frac{{OP}}{{OQ}}\,\,or\,\frac{{OP}}{{O{P_1}}} \times \frac{{ON}}{{O{Q_1}}}\)

\(\frac{{OP}}{{O{P_1}}}\) is common and therefore, \({e_d}\) can be compared by comparing \(\frac{{OM}}{{OQ}}\,and\frac{{ON}}{{O{Q_1}}}\).

\(But\,\frac{{OM}}{{OQ}}\, = \frac{{ON}}{{O{Q_1}}}\).

This can be proved geometrically