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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

Verified by Experts

We have `tanalpha=|(m-m)/(1+mm')|'` where
m and m' are the slopes of
`3x-4y+1=0 and 4x+y=1` respectively
`therefore m=3/4m==-4`

`thereforetanalpha=|((3)/(4)+4)/(1-3)|=19/8`
`therefore` The equation of AC is
either `y+7=tan (theta-alpha)(x-2)" "...(i)`
or` y+7tan(theta+alpha)(x-2)" "...(ii)`
where `tan theta=m=3/4`
Now, `tan(theta-alpha)=((3)/(4)-(19)/(8))/(1+(57)/(32))=-52/89and tan(theta+alpha)=((3)/(4)+(19)/(8))/(1-(57)/(32))=-4`
which is the same as slope of AB.
`therefore` equation of the required line is
`y+7=(-52)/(89)(x-2)or52x+89y+519=0`
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