A straight line $\frac{x}{a}-\frac{y}{b}=1$ passes through the point $(8,6)$ and cuts a triangle of area 12 sq units from the axes of coordinates. The equations of the straight lines are

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Aicharya · Accepted Answer

A straight line $\frac{x}{a}-\frac{y}{b}=1$ passes thought $(8,6)$. Then, $\frac{8}{a}-\frac{6}{b}=1...(i)$ Area of triangle formed by axis and line $=12 \mathrm{sq}$ unit

$ \begin{aligned} & \text { Area of triangle }=\frac{1}{2} \times \text { base } \times \text { height } \\ & \qquad \begin{aligned} 12 & =\frac{1}{2}|(a) \times(-b)| \Rightarrow 24=|-a b| \\ |a b|=24, a b & = \pm 24, b=\frac{ \pm 24}{a} \end{aligned} \end{aligned} $ Now, in Eq. (i) put $b=24 / a$ $\begin{array}{rlrl} & & \frac{8}{a}-\frac{6}{b} & =1 \\ \Rightarrow & & \frac{8}{a}-\frac{6}{\frac{24}{a}} & =1 \Rightarrow \frac{8}{a}-\frac{a}{4}=1 \\ \Rightarrow & & \frac{32-a^2}{4 a} & =1 \Rightarrow 32-a^2=4 a \\ \Rightarrow & a^2+4 a-32 & =0 \Rightarrow(a+8)(a-4)=0 \\ & a=-8 \mid a & =4 \\ & b=-3 \mid b & =6\end{array}$ $\therefore$ Equation of lines are \begin{array}{c|l} \frac{x}{-8}-\frac{y}{-3}=1 & \frac{x}{4}-\frac{y}{6}=1 \\ -3 x+8 y=24 & \frac{3 x-2 y}{12}=1 \\ 3 x-8 y+24=0 & \begin{array}{l}3 x-2 y=12 \\ 3 x-2 y-12=0\end{array} \end{array}

A straight line $\frac{x}{a}-\frac{y}{b}=1$ passes through the point $(8,6)$ and cuts a triangle of area 12…

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