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Mathematics Question on Straight lines

The area (in square unit) of the triangle formed by x+y+1=0x+y+1=0 and the pair of straight lines x2−3xy=2y2=0x^{2}-3xy=2y^{2}=0 is

A

712\frac{7}{12}

B

512\frac{5}{12}

C

112\frac{1}{12}

D

16\frac{1}{6}

Answer

112\frac{1}{12}

Explanation

Solution

x2−2xy−xy+2y2=0x^{2}-2xy-xy+2y^{2}=0
⇒(x−2y)(x−y)=0\Rightarrow \left(x-2y\right)\left(x-y\right)=0
⇒x=2y,x=y=0   …(i)\Rightarrow x=2y, x=y=0\,\,\,\ldots\left(i\right)
Also, x+y+1=0    …(ii)x+y+1=0\,\,\,\,\ldots\left(ii\right)
On solving Eqs. (i) and (ii), we get
A(−23,−13),B(−12,−12),C(0,0)A\left(-\frac{2}{3}, -\frac{1}{3}\right), B\left(-\frac{1}{2},-\frac{1}{2}\right),C\left(0, 0\right)
∴\therefore Area of ΔABC=12∣23−131 12−121 001∣\Delta ABC=\frac{1}{2}\begin{vmatrix}\frac{2}{3}&-\frac{1}{3}&1\\\ \frac{1}{2}&-\frac{1}{2}&1\\\ 0&0&1\end{vmatrix}
=12[13−16]=12[16]=112=\frac{1}{2}\left[\frac{1}{3}-\frac{1}{6}\right]=\frac{1}{2}\left[\frac{1}{6}\right]=\frac{1}{12}