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

Question

Aicharya · Accepted Answer

Since, intercept form of line is $\frac{x}{a}+\frac{y}{b}=1$ where X-intercept = a, Y-intercept = b

Hence, equation of line AB $ \frac{x}{m}+\frac{y}{n}=1 $ $\because$ Line (i) passes through $(4,5)$. $ \begin{aligned} \therefore \quad \frac{4}{m} & +\frac{5}{n}=1 \\ \frac{4}{m} & =1-\frac{5}{n} \\ m & =\frac{4 n}{n-5} \end{aligned} $ $\begin{gathered}\because \text { Area of } \triangle O A B=\frac{1}{2} \times O A \times O B \\ A=\frac{1}{2} m n=\frac{1}{2}\left(\frac{4 n}{n-5}\right) n \\ A=\frac{2 n^2}{n-5} \\ \therefore \frac{d A}{d n}=2\left\{\frac{(n-5) \cdot 2 n-n^2(1)}{(n-5)^2}\right\} \\ \frac{d A}{d n}=2\left\{\frac{2 n^2-10 n-n^2}{(n-5)^2}\right\} \\ =\frac{2 n^2-20 n}{(n-5)^2}\end{gathered}$ $\because$ Area is least for some value of $n$. $ \begin{array}{ll} \therefore \quad & \frac{d A}{d n}=0 \\ & 2 n^2-20 n=0 \\ & 2 n(n-10)=0 \\ \Rightarrow \quad \quad \quad n=0,10 \\ \text { At } & n=10, m=\frac{4 n}{n-5}=\frac{4 \times 10}{10-5}=\frac{40}{5} \\ \quad & m=8 \\ \therefore m & n=8: 10=4: 5 \end{array} $

Through the point (4, 5), a straight line is drawn making positive intercepts on the coordinate axes. The…

Solution