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A line 4x+y=1 passes through the point A...

A line `4x+y=1` passes through the point A(2,-7) and meets line BC at B whose equation is `3x -4y +1=0`, the equation of line AC such that `AB=AC` is (a) 52x +89y +519=0(b) 52x +89y-519=0 c) 82x +52y+519=0 (d) 89x +52y -519=0

Text Solution

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Let the equation of BC be
`3x-4y+1=0 " " (1)`
and the equation of AB be
`4x+y-1 = 0 " "(2)`
Since AB=AC, we have
`angle ABC = angle ACB = alpha ("say")`
` "Slope of line BC" = (3)/(4)`
and slope of AB =-4
Let the slope of a AC be m. Equating the two values of tan a, we get
`|(-4-(3)/(4))/(1-4 xx (3)/(4))| = | ((3)/(4)-m)/(1+(3)/(4)m)|`
`" or " +-(19)/(8) = (3-4m)/(4+3m)`
`" or " m = (52)/(89)-4`
Therefore, the equation of AC is
`y+7 = -((52)/(89))(x-2) " or " 52x + 89y + 519 = 0`
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