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Question

Q Find the equation of the line passing through the intersection of the lines 4x+3y=1 and 5x+4y=2 and
i) parallel to the line x+2y-5=0
ii) perpendicular to x-axis.

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Solution

Dear student
The intersection of the line 4x+3y=1 ...1and5x+4y=2 ...2can be find by solving the two equations simultaneously.Multiply 1 by 5 and 2 by 4.20x+15y=5 ...320x+16y=8 ...4Subtracting 3 from 4, we get20x+16y-20x-15y=8-5y=3Putting in 1, we get4x+33=14x=-8x=-2Thus, the point of intersection of two lines is -2,3This point lies on the required line.i Slope of line y=mx+c is m.The slope of the required line is same as that of line when two lines are , their slopes are equal.x+2y-5=02y=-x+5y=-12x+52Thus, the slope of line is -12So, equation of the required line isy-y1=mx-x1where m is the slope and x1,y1= -2,3 is a point on line.Thus, equation of required line is:y-3=-12x--2y-3=-12x+22y-6=-x-2x+2y-4=0ii to x axis.Equation of line to x axis is given by x=k, where k is a constant-2,3 lies on the line x=k givenSo, k=-2Thus, the equation of straight line is x=-2 or x+2=0
Regards

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