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Question
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 that point A also lies on the line 3x + k( y + 1 ) = 0, find the value of k.
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Solution
Let the coordinates of A be`(x,y) Here (PA)/(PQ) = 2/5 . so ,`
PA + AQ= PQ
`⇒PA +AQ =(5PA)/2 [∵ PA = 2/5 PQ]`
` ⇒AQ = (5PA)/2 - PA`
`⇒ (AQ)/(PA) = 3/2 `
`⇒ (PA)/(AQ) = 2/3 `
Let (x, y) be the coordinates of A, which dives PQ in the ratio 2 : 3 internally Then using section formula, we get
` X = (2 xx (-4) +3 xx (6))/(2+3) = (-8+18)/5= 10/5 = 2`
`y = (2 xx (-1) + 3 xx(-6))/(2+3) = (-2-18)/5 = (-20)/5 = -4`
Now, the point ( 2, -4 ) lies on the line 3x +k(y+1) = 0 ,therefore
3 × 2 +k(-4+1)=0
⇒ 3k = 6
`⇒ k =6/3 =2`
Hence, k=2.
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