point r divides the line segment joining the points p(3,2) and (6,-7) such that pr/pq=1/3 if r lies on the line 3x-4y+k=o find the value of k
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we have pr:pq= 1:3
Now applying section formula
r(x,y) = [ 1(6)+3(3)]/[1+3] , [1(-7)+3(2)]/[1+3]
= (6+9)/4 , (-7+6)/4
= 15/4,-1/4
Thus x = 15/4
and y = -1/4
Putting values of x and y in eq. 3x-4y+k=0
3(15/4)-4(-1/4)+k=0
= k=-49/4
Now applying section formula
r(x,y) = [ 1(6)+3(3)]/[1+3] , [1(-7)+3(2)]/[1+3]
= (6+9)/4 , (-7+6)/4
= 15/4,-1/4
Thus x = 15/4
and y = -1/4
Putting values of x and y in eq. 3x-4y+k=0
3(15/4)-4(-1/4)+k=0
= k=-49/4
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