A straight line through the point `A(-2, -3)` cuts the line `x+3y=9` and `x+y+1=0` at `B and C` respectively. Find the equation of the line if `AB.AC = 20`.

A straight line through the point A (-2,-3) cuts the line x+3y=9 and x+y+1=0 at B and C respectively. If AB.AC =20 then equation of the possible line is

A straight line through the point A (-2,-3) cuts the line x+3y=9 and x+y+1=0 at B and C respectively. If AB.AC =20 then equation of the possible line is

A straight line through the point A (-2,-3) cuts the line x+3y=9 and x+y+1=0 at B and C respectively. If AB.AC =20 then equation of the possible line is

A straight line through the point (2,2) intersects the lines sqrt(3)x+y=0 and sqrt(3)x-y=0 at the point A and B respectively.Then find the equation of the line AB so that triangle OAB is equilateral.

A straight line L passing through the point P(-2,-3) cuts the lines, x + 3y = 9 and x+y+1=0 at Q and R respectively. If (PQ) (PR) = 20 then the slope line L can be (A) 4 (B) 1 (C) 2 (D) 3

A line through A(-5,-4) meets the lines x+3y+2=0, 2x+y+4= and x-y-5=0 at the points B,C,D respectively. Find the equation of the line if (15/(AB))^(2)+(10/(AC))^(2)=(6/(AD))^(2)

A straight line through the point (2, 2) intersects the lines sqrt(3)x+y=0 and sqrt(3)x-y=0 at the points A and B. The equation of the line AB so that DeltaOAB is equilateral, is:

A straight line through the point (2, 2) intersects the lines sqrt3x + y = 0 and sqrt3 x-y=0 at the points A and B. The equation to the line AB so that the triangle OAB is equilateral is ,