Question

Find the equation of the line passing through the intersection of $2x+y=8,3x–7=y$ and parallel to $4x+y=11$.

• A. $4x+y=14$
• B. $x+4y-10=0$
• C. $3x+4y-14=0$
• D. $3x+y-11=0$

Answer 1

The correct option is A

$4x+y=14$

To find the point of intersection, we need to solve the equations

$2x+y=8$ and $3x–7=y$

$2x+y=8$...... (1)

$3x–7=y$........(2)

Substituting the value of $y$ from eq(2) to eq(1), we get

$\Rightarrow 2x+y=8$

$\Rightarrow 2x+3x-7=8$

$\Rightarrow 5x-7=8$

$\Rightarrow 5x=15$

$\Rightarrow x=3$

Substituting $x=3$ in eq(2), we get

$\Rightarrow y=3x–7$

$\Rightarrow y=3\left(3\right)–7$

$\Rightarrow y=9–7$

$\Rightarrow y=2$

$\therefore x=3,y=2$

The line $4x+y=11$ can be written in the form of $y=-4x+11$.

Comparing $y=-4x+11$ with $y=mx+c$, we get

Slope, m = -4.

Therefore the line parallel to $4x+y=11$ will have a slope of -4.

So, equation of the required line is $(y-2)=(-4)(x-3)$

$y-2=-4x+12$

$4x+y=14$