1.

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

Answer»

`x - y=1`
`x - y +1 = 0`
`3x -y +3 = 0`
`3x -y = 3`

Solution :Any point on line through A is
`(-2 +r cos theta, -3 +r sin theta)`
`:. (-2+AB cos theta, -3 +AB sin theta)` lies on `x +3Y = 9`
`:. AB = (20)/((cos theta +3 sin theta))`, similarly `AC = (4)/((cos theta + sin theta))`
`AB xx AC = 20`
`:. 4 = cos^(2) theta +4 sin theta cos theta +3 sin^(2) theta`
`:. 4 +4 tan^(2) theta = 1 +4 tan theta +3 tan^(2) theta`
`:. tan^(2) theta - 4 tan theta +3 = 0`
`:. tan theta = 1` or `tan theta = 3`
`:.` Required lines are
`y +3 =x +2` or `y +3 =3 (x+2)`


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