Slope between two lines

Geometry Level 1

If the line x + a y + 1 = 0 x+ay+1=0 is perpendicular to the line 2 x − b y + 1 = 0 2x-by+1=0 , and parallel to the line x − ( b − 3 ) y − 1 = 0 , x-(b-3)y-1=0, what is the value of a + b ? a+b?

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Sandeep Mehra by Sandeep Mehra , 23, India

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2 solutions

Consider linear equation y=mx+c. The slope of the line is m. So the slope m1= -1/a, the slope m2= 2/b and the slope m3= -1/(b-3). Since line1 parallel to line3, we have slope m1=m3. Then -1/a=-1/(b-3). Rearrange, we have a+b=3.

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Mijeeb Jim
Nov 28, 2014

The line x+ay+1=0 is perpendicular to the second line 2x-by+1=0 so their slopes are negative reciprocal to each other, so the slope of the first line is m1= -1/a and slope for the second line is m2 = 2/b then we equate them m1 = -1 / m2 or m2=-1/m1 so we get a = 2/b , for the third line which is x-(b-3)y-1=0 is parallel to the first line thus their slopes are equal , m3 = 1 / (b-3) then equate m1 = m3 thus we get a = -b + 3 , then substitute the equation a = 2/b after that we get a quadratic equation which is b2 -3b+2=0 then apply factoring we get b = 1 or b =2 , if we use b=1 we get a =2 or if we use b = 2 we get a =1. Thus the question a+b = 3

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