Question

The equation of the line cutting off an intercept of 3 on the Y-axis and parallel to the line joining points A(4, -5) and B(1, 2) is .

• A. 3x+7y-9 = 0
• B. 3x-7y+9 = 0
• C. 3y-7x+9 = 0
• D. 3y+7x-9 = 0

Answer 1

The correct option is D 3y+7x-9 = 0

${\mathrm{Slope}}_{\mathrm{AB}}=\frac{2-(-5)}{1-4}=\frac{-7}{3}\mathrm{Since}\mathrm{parallel}\mathrm{lines}\mathrm{have}\mathrm{same}\mathrm{slope},\mathrm{Slope}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{line}=\frac{-7}{3}\mathrm{Given}Y-\mathrm{intercept}=3\therefore \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{line}\mathrm{is}y=\frac{-7}{3}x+3\Rightarrow 3y=-7x+9\Rightarrow 3y+7x-9=0$