The line 3x-4y+7=0 is rotated through an angle (π /4) in the clockwise direction about the point (- 1 , 1). The equation of the line in its new position is

Question

The line 3x-4y+7=0 is rotated through an angle (π /4) in the clockwise direction about the point (- 1 , 1). The equation of the line in its new position is (A) 7y+x-6=0 (B) 7y-x-6=0 (C) 7y+x+6=0 (D) 7y-x+6=0

Tardigrade Admin · Accepted Answer

As (.-1,1.) is a point on 3x - 4y + 7 = 0, the rotation is possible. Slope of the given line =(3/4) Slope of the line in its new position =((3/4) - 1/1 + (3)4)=-(1/7) Therefore, the required equation is y-1=-(1/7) (x + 1) or 7y+x-6=0

Q. The line $3 x - 4 y + 7 = 0$ is rotated through an angle $\frac{π}{4}$ in the clockwise direction about the point $(- 1, 1) .$ The equation of the line in its new position is

$7 y + x - 6 = 0$

$7 y - x - 6 = 0$

$7 y + x + 6 = 0$

$7 y - x + 6 = 0$

Q. The line 3x−4y+7=0 is rotated through an angle 4π​ in the clockwise direction about the point (−1,1). The equation of the line in its new position is

7y+x−6=0

7y−x−6=0

7y+x+6=0

7y−x+6=0

As (−1,1) is a point on 3x−4y+7=0, the rotation is possible. Slope of the given line =43​ Slope of the line in its new position =1+43​43​−1​=−71​ Therefore, the required equation is y−1=−71​(x+1) or 7y+x−6=0

Q. The line $3 x - 4 y + 7 = 0$ is rotated through an angle $\frac{π}{4}$ in the clockwise direction about the point $(- 1, 1) .$ The equation of the line in its new position is

$7 y + x - 6 = 0$

$7 y - x - 6 = 0$

$7 y + x + 6 = 0$

$7 y - x + 6 = 0$