Question

If $P(x,y)$ be a variable point on the line $y=2x$ lying between the lines $2(x+1)+y=0$ and $x+3(y-1)=0$, then

• A. $x\in (-\frac{1}{2},\frac{6}{7})$
• B. $x\in (-\frac{1}{2},\frac{3}{7})$
• C. $y\in (-1,\frac{3}{7})$
• D. $y\in (-1,\frac{6}{7})$

Answer 1

Answer: (B), (D)

The correct options are
B $x\in (-\frac{1}{2},\frac{3}{7})$
D $y\in (-1,\frac{6}{7})$

Substituting $y=2x$ in the two given equations $2x+y+2=0$ and $x+3y-3=0$
We get $4x+2=0=>x=-\frac{1}{2}$ and $-1+y+2=0=>y=-1$
Other line $x+3\left(2x\right)-3=0=>7x-3=0=>x=\frac{3}{7}$ and $y=\frac{6}{7}$
This values of $x,y$ give the range
Therefore $x\in (\frac{-1}{2},\frac{3}{7})$ and $y\in (-1,\frac{6}{7})$