Question

Equation of the straight line passing through the point of intersection of the lines $3x+4y=7,x-y+2=0$ and having slope $3$ is

• A. 21x−7y+16=0
• B. 9x−3y+14=0
• C. 21x−7y+12=0
• D. 9x−3y+5=0

Answer 1

The correct option is A

21x−7y+16=0

On solving the two equations, we get the point of intersection as ($\frac{-1}{7}$, $\frac{13}{7}$)

Slope of the required equation $m=3$

Therefore, using point-slope form

$(y-y_1)=m(x-x_1)$

$y-\frac{13}{7}=3(x-\frac{-1}{7})$

$7y-13=21x+3$

$21x-7y+16=0$