The area of the triangle formed by the coordi | vedclass

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(A) The line equation is $px + qy = r$. Dividing by $r$, we get $\frac{x}{r/p} + \frac{y}{r/q} = 1$.The intercepts on the axes are $a = \frac{r}{p}$ a

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$p, 3r, q$ are in $G$.$P$.

Vedclass · Answer

(A) The line equation is $px + qy = r$. Dividing by $r$, we get $\frac{x}{r/p} + \frac{y}{r/q} = 1$.The intercepts on the axes are $a = \frac{r}{p}$ and $b = \frac{r}{q}$.The area of the triangle formed is $\frac{1}{2} 	imes |a| 	imes |b| = \frac{1}{2} 	imes \frac{r}{p} 	imes \frac{r}{q} = \frac{r^2}{2pq}$.Given area is $\frac{1}{54}$, so $\frac{r^2}{2pq} = \frac{1}{54}$.This simplifies to $2pq = 54r^2$, or $pq = 27r^2$.Since $p, 27r, q$ are in $G$.$P$., the middle term squared equals the product of the extremes: $(27r)^2 = p 	imes q$, which is $729r^2 = pq$. This does not match.Wait, let us re-evaluate: $\frac{r^2}{2pq} = \frac{1}{54} \implies pq = 27r^2$. If $p, k, q$ are in $G$.$P$., then $k^2 = pq$. Here $k^2 = 27r^2$, so $k = \sqrt{27}r = 3\sqrt{3}r$. The options provided suggest a typo in the problem statement or options. Assuming the area was $\frac{r^2}{2pq} = \frac{1}{2pq}$, the relation is $pq = r^2$. Given the options, if $pq = (3r)^2 = 9r^2$, then area is $\frac{r^2}{18r^2} = \frac{1}{18}$. If $pq = (27r)^2$, area is $\frac{1}{1458}$. Given the standard form, the correct relation is $pq = 27r^2$.

The area of the triangle formed by the coordinate axes and the line $px + qy = r$ (where $p, q$ and $r$ are positive real numbers) is $\frac{r^2}{2pq}$. If this area is $\frac{1}{54}$ sq. units, then which of the following is true?

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