If the area of the triangle formed by the pair of lines $8 x^2-6 x y+y^2=0$ and the line $2 x+3 y=a$ is 7 , then $a$ is equal to

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Aicharya · Accepted Answer

Given pair of lines is $8 x^2-6 x y+y^2=0$. $\begin{aligned} & \Rightarrow \quad 8\left(\frac{x}{y}ight)^2-6\left(\frac{x}{y}ight)+1=0 \ & \Rightarrow \quad\left[4\left(\frac{x}{y}ight)-1ight]\left[2\left(\frac{x}{y}ight)-1ight]=0 \ & \Rightarrow \quad \frac{4 x}{y}=1 	ext { and } \frac{2 x}{y}=1 \ & \Rightarrow \quad 4 x=y \quad 	ext { and } \quad 2 x=y \ & \end{aligned}$ Also, other line is $2 x+3 y=a$ On solving Eqs. (i) and (ii), we get $O(0,0), A\left(\frac{a}{14}, \frac{2 a}{7}ight)$ and $B\left(\frac{a}{8}, \frac{a}{4}ight)$ $\begin{array}{ll}	herefore 	ext { Area of } 	riangle A B C=\frac{1}{2}|| \begin{array}{ccc}0 & 0 & 1 \ \frac{a}{14} & \frac{2 a}{7} & 1 \ \frac{a}{8} & \frac{a}{4} & 1\end{array} \| \ \Rightarrow & 7=\frac{1}{2}\left|\left(\frac{a^2}{56}-\frac{2 a^2}{56}ight)ight| \ \Rightarrow & 7=\frac{1}{2}\left|\frac{-a^2}{56}ight| \ \Rightarrow & 14 	imes 56=a^2 \ \Rightarrow & a=28\end{array}$

If the area of the triangle formed by the pair of lines $8 x^2-6 x y+y^2=0$ and the line $2 x+3 y=a$ is 7 ,…

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