Mathematics  Locus of a Point
Mathematics

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Find the equation(s) of the locus of P that moves such that the distance from P to the lines 3x 4y + 1 = 0 and 12x + 5y + 3 = 0 is in the ratio 3:1
I am glad I found you again!
Sorry dude, my answer cut off in the middle and, so, I gave you only half of the answer first time. Just realized it! That’s why it looks so weird at the end first time!
Here is the full answer:
3x  4y + 1 = 0
12x + 5y + 3 = 0
Simplify, Find x from the first equation and substitute x with the final result in a second equation to find y within the next steps:
1)
3x = 4y  1, x = (4 y 1) / 3
5y =  3  12x, y =  (3 + 12x) / 5
2)
x = (4 ((3+12x)/5 )  1) / 3
x = ((( 12  48x) / 5)  1) /3
x = (( 12  48x  5) / 5) /3
x = (( 17  48x)/5 ) /3
x = 17/15  48x/15 = (17  48x)/15
x  ((17+48x)/15) = 0
x/15  (17+48x)/15 = 0
(15x + 48x + 17)/15 = 0
(63x + 17)/15 =0
63x = 17
x =  17/63
3)
y = ( (3 + 12( 17/63))) /5
y =  (( 3 – 204/63) )/5
y =  ((189  204)/63)/5
y = (15/63)/5
y = 1/21
4)
D= │ax + by + c│ / √a ^{2} + b^{2}
You can take any of those condition equations in your question. Let’s take first:
3x  4y + 1 = 0
Here, a = 3; b=4; c = 1
Substitute and plug them into equation for D in a ration 1:3 as stated:
1/3 = │3x17/63 + 4x1/21 + C│/ 5
5/3 = │17/21 + 4/21 + C│
5/3 = 21/21 + C
5/3 – C = + 1
C1 = 5/3 + 1 = 8/3
C2 = 5/3 – 1 = 2/3
2 is 3 times smaller than 8, corresponding to the 1:3 ratio
Having C1 and C2 in chosen equation, you will end up with two equations as:
3x  4y + 8/3 = 0
3x  4y + 2/3 = 0
Hope this answer found and got you on time!
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