The line joining the points (2,1) and (5,8) is trisected by the points P and Q.If the point P lies on line 2x-y+K=0 ,find the value of K ?_____________solve in two to three steps if not solve long?

1 Answer
CW
Oct 1, 2017

#K=-8/3#

Explanation:

Section formula :
If a point #P(x,y)# divides a line segment joining #A(x_1,y_1)and B(x_2,y_2)# in the ratio of #m:n#, i.e., #(AP:PB=m:n)#,
then #P(x,y)= ((mx_2+nx_1)/(m+n), (my_2+ny_1)/(m+n))#
Given the line joining #A(2,1) and B(5,8)# is trisected by #P and Q#,
#=> AP:PQ:QB=1:1:1#,
#=> AP:PB=1:2#
#=> P(x,y)=((1xx5+2xx2)/(1+2),(1xx8+2xx1)/(1+2))=(3,10/3)#
Given #P# lies on the line #2x-y+K=0#,
#=> 2(3)-10/3+K=0#
#=> K=10/3-6=10/3-(18)/3=-8/3#

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