Analytical Geometry
find the value of k so that straight line 2x+3y+4+k(6x-y+12)= 0 is perpendicular to the line 7x+5y-4 =0
find the value of k so that straight line 2x+3y+4+k(6x-y+12)= 0 is perpendicular to the line 7x+5y-4 =0
1 Answer
xyz xz
2x+3y+4+k(6x-y+12) = 0
=x(2-6k) + y(3-k) + (4+12k)=0
slope is (6k-2)/(3-k)
but this line is perpendicular to 7x+5y-4=0 and its slope is -7/5
imples
(-7/5)(6k-2)/(3-k) = -1
implies k = 29/47
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