Point A lies on the line segment PQ joining P[6,-6] and Q [-4,-1] in such a way that PA / PQ =2/5 . If the point A lies on the line 3x+k[y+1]=0 ,find the value of k. solve this one only .
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To approach this question, you can use the distance of PQ and the PA / PQ =2/5 to find point A.
Using point A, you can find k in the line.
Full solution is attached.
Using point A, you can find k in the line.
Full solution is attached.
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Answered by
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since A,P,Q lies on the same line segment
then using section formula
point P will divide AQ inthe ratio of 2:5
PA:PQ=2:5
let the cordinates of point A(x,y)
6={(2*(-4))-5x}/(2-5)
6={-8-5x}/(-3)
18=8+5x
x=2
similarly => -6={2*(-1)-5y}/(2-5)
-6={-2-5y}/(-3)
-18=2+5y
y=-4
A(2,-4)
putting in the given equation
3x+k(y+1)=0
3*2 + k(-4+1)=0
6-3k=0
k=2
then using section formula
point P will divide AQ inthe ratio of 2:5
PA:PQ=2:5
let the cordinates of point A(x,y)
6={(2*(-4))-5x}/(2-5)
6={-8-5x}/(-3)
18=8+5x
x=2
similarly => -6={2*(-1)-5y}/(2-5)
-6={-2-5y}/(-3)
-18=2+5y
y=-4
A(2,-4)
putting in the given equation
3x+k(y+1)=0
3*2 + k(-4+1)=0
6-3k=0
k=2
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