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 (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.

Two straight lines passing through the point A(3, 2) cut the line 2y=x+3 and x-axis perpendicularly at P and Q respectively. The equation of the line PQ is

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 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 P(-2,-3) cuts the pair of straight lines x^(2)+3y^(2)+4xy-8x-6y-9=0 in Q and R Find the equation of the line if PQ.PR=20

A straight line through origin O meets the lines 3y=10-4x and 8x+6y+5=0 at point A and B respectively. Then , O divides the degment AB in the ratio.