The straight line y = x - 2 rotates about a point where it cuts the x-axis and becomes perpendicular to the straight line ax + by + c = 0. Then its equation is

Question

The straight line y = x - 2 rotates about a point where it cuts the x-axis and becomes perpendicular to the straight line ax + by + c = 0. Then its equation is (A) ax + by + 2a = 0 (B) ax - by - 2a = 0 (C) bx + ay - 2b = 0 (D) ay - bx + 2b = 0

Tardigrade Admin · Accepted Answer

Slope of the line in the new position is

\frac{b}{a}

, since it is

⊥

to the line

a x + b y + c = 0

and it cuts the x-axis at

(2, 0)

. Hence, the required line passes through

(2, 0)

and its slope is

\frac{b}{a}

. Required eq. is

y - 0 = \frac{b}{a} (x - 2) \Rightarrow a y = b x - 2 b \Rightarrow a y - b x + 2 b = 0

Q. The straight line $y = x - 2$ rotates about a point where it cuts the x-axis and becomes perpendicular to the straight line $a x + b y + c = 0$ . Then its equation is

$a x + b y + 2 a = 0$

$a x - b y - 2 a = 0$

$b x + a y - 2 b = 0$

$a y - b x + 2 b = 0$

Q. The straight line y=x−2 rotates about a point where it cuts the x-axis and becomes perpendicular to the straight line ax+by+c=0. Then its equation is

ax+by+2a=0

ax−by−2a=0

bx+ay−2b=0

ay−bx+2b=0

Slope of the line in the new position is ab​, since it is ⊥ to the line ax+by+c=0 and it cuts the x-axis at (2,0). Hence, the required line passes through (2,0) and its slope is ab​. Required eq. is y−0=ab​(x−2)⇒ay=bx−2b⇒ay−bx+2b=0

Q. The straight line $y = x - 2$ rotates about a point where it cuts the x-axis and becomes perpendicular to the straight line $a x + b y + c = 0$ . Then its equation is

$a x + b y + 2 a = 0$

$a x - b y - 2 a = 0$

$b x + a y - 2 b = 0$

$a y - b x + 2 b = 0$