KeerthanaPosted on What is the equation of a line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3? ax + 7y = 21 b7x + y = 21 cx + 2y = 10 d–x + y = 15 Answer : Option BExplanation : Let the slope of line be m Here, m1=−1−7=17m_{1}=\frac{-1}{-7}=\frac{1}{7}m1=−7−1=71 As lines are perpendicular, ∴m1×m2=−1\therefore m_{1} \times m_{2}=-1∴m1×m2=−1 m×17=−1m \times \frac{1}{7}=-1m×71=−1 m=−7m=-7m=−7 ∴\therefore∴ Equation of line is (y−y1)=m(x−x1)\left(y-y_{1}\right)=m\left(x-x_{1}\right)(y−y1)=m(x−x1) (y−0)=−7(x−3)(y-0)=-7(x-3)(y−0)=−7(x−3) y=−7x+21y=-7 x+21y=−7x+21 ⇒7x+y=21\Rightarrow 7 x+y=21⇒7x+y=21 Rate This:NaN / 5 - 1 votesAdd comment