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Question

The equation of the straight line passing through the point of intersection of x+2y=5 and 3x+7y=17 and perpendicular to the straight line 3x+4y=10 is

A
4x3y+1=0
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B
4x3y+12=0
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C
4x3y2=0
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D
4x3y+2=0
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Solution

The correct option is D 4x3y+2=0
Equation of line passing through the intersection of the lines x+2y5=0 and 3x+7y17=0 is

(x+2y5)+λ(3x+7y17)=0
x(3λ+1)+y(7λ+2)(17λ+5)=0(1)
Slope of the line
m1=(3λ+17λ+2)
Now, slope of the other given line
3x+4y10=0m2=34

As the lines are perpendicular, so
m1m2=1(3λ+17λ+2)(34)=19λ+3=28λ8λ=1137

Putting the value of λ in equation (1), we get
x(3337+1)+y(7737+2)(18737+5)=0
4x3y+2=0


Alternate solution:
Given line 3x+4y10=0
Perpendicular line to this
4x3y+μ=0(2)
Now finding the point of intersection of
x+2y5=0 and 3x+7y17=0
We get
x=52y3(52y)+7y17=0y=2x=1
Using equation (2), we get
46+μ=0μ=2
So, the equation of the required line is
4x3y+2=0

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