Through the point (4, 5),a straight line is d | vedclass

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(C) Let the intercepts on the $X$ and $Y$ axes be $m$ and $n$ respectively. The equation of the line is $\frac{x}{m} + \frac{y}{n} = 1$.Since the line

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$4 : 5$

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(C) Let the intercepts on the $X$ and $Y$ axes be $m$ and $n$ respectively. The equation of the line is $\frac{x}{m} + \frac{y}{n} = 1$.Since the line passes through $(4, 5)$,we have $\frac{4}{m} + \frac{5}{n} = 1$,which implies $\frac{4}{m} = 1 - \frac{5}{n} = \frac{n-5}{n}$,so $m = \frac{4n}{n-5}$.The area of the triangle formed with the coordinate axes is $A = \frac{1}{2}mn = \frac{1}{2} \left( \frac{4n}{n-5} \right) n = \frac{2n^2}{n-5}$.To find the minimum area,we differentiate $A$ with respect to $n$:$\frac{dA}{dn} = 2 \left[ \frac{(n-5)(2n) - n^2(1)}{(n-5)^2} \right] = 2 \left[ \frac{2n^2 - 10n - n^2}{(n-5)^2} \right] = \frac{2n^2 - 20n}{(n-5)^2}$.Setting $\frac{dA}{dn} = 0$,we get $2n(n-10) = 0$. Since $n > 5$ for positive intercepts,we have $n = 10$.Then $m = \frac{4(10)}{10-5} = \frac{40}{5} = 8$.The ratio of the intercepts $m : n = 8 : 10 = 4 : 5$.

Through the point $(4, 5)$,a straight line is drawn making positive intercepts on the coordinate axes. The area of the triangle thus formed is least,when the ratio of the intercepts on the $X$ and $Y$ axes is

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