. If the distances from P to the points (5,–4), (7, 6) are in the ratio 2 : 3, then find the locus
of P
Answers
Answer:
5x^ 2 +5y^2 −34x+120y+29=0
Step-by-step explanation:
Let P(x,y) be a point in locus.
The given points are A(5,−4) and B(7,6) and given that PA:PB=2:3. Therefore, we have:
PA /PB = 3/2
⇒3PA=2PB
⇒9PA ^2 =4PB ^2
Now, applying the distance formula to the given points A(5,−4) and B(7,6):
9[(x−5)^2 +(y+4) ^2 ]=4[(x−7) ^2 +(y−6) ^2 ]
⇒9[x ^2 +5 ^2 −(2×x×5)+y ^2 +4 ^2 +(2×y×4)]=4[x ^2 +7 ^2 −(2×x×7)+y ^2 +6 ^2 −(2×y×6)] (∵(a+b) ^2) =a ^2 +b ^2 +2ab,(a−b) ^2 =a ^2 +b^2 −2ab)
⇒9(x ^2 +25−10x+y ^2 +16+8y)=4(x ^2 +49−14x+y ^2 +36−12y)
⇒9(x ^2 +y^2 −10x+8y+41)=4(x ^2+y^2 −14x−12y+85)
⇒9x ^2 +9y ^2 −90x+72y+369=4x ^2 +4y ^2 −56x−48y+340
⇒9x^2 −4x ^2 +9y ^2 −4y ^2−90x+56x+72y+48y+369−340=0
⇒5x ^2 +5y^ 2 −34x+120y+29=0
Hence, the equation of locus of P is 5x^ 2 +5y^2 −34x+120y+29=0