Question

The angle between the lines represented by $2x^2+5xy+6x+7y+4=0$ is ${\tan }^{-1}\left(m\right)$ and $a^2+b^2-ab-a-b+1\le 0$, then $2a+3b=$

• A. $\frac{1}{m}$
• B. $m$
• C. $-m$
• D. $m^2$

Answer 1

The correct option is A $\frac{1}{m}$

$\begin{array}{c}\text{When that,}\tan \theta =\frac{2\sqrt{h^2-ab}}{|a+b|}\text{for}\\ a{x}^2+b{y}^2+2hxy+2gx+2fy+c=0\\ \Rightarrow \tan \theta =\frac{2\sqrt{\frac{25}{4}-6}}{151}=\frac{1}{5}=m\\ \text{Also,}{a}^2+b^2-ab-a-b+1⩽0\\ \text{then}2a+3b=2\times 2+(-3)\times 3=4-9=n-5\\ \Rightarrow 2a+3b=\frac{-1}{m}\end{array}$