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Can anyone solve these locus problems? (1 Viewer)

Jmmalic220

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Can anyone please help me solve these problems?

1. Find the equation of the locus of a point that moves so that it is equidistant from the line 4x-3y+2=0 and the line 3x+4y-7=0.

2. Given two points A(3, -2) and B(-1, 7), find the equation of the locus of P(x, y) if the gradient of PA is twice the gradient of PB.

3. If R is the fixed point (3,2) and P is a movable point (x,y), find the equation of the locus of P if the distance PR is twice the distance from P to the line y=-1.

I have the answers but I need to understand how to do it. Please reply with workings.

Thank you very much in advance!
 
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davidgoes4wce

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Can anyone please help me solve these problems?

1. Find the equation of the locus of a point that moves so that it is equidistant from the line 4x-3y+2=0 and the line 3x+4y-7=0.









1 Where both sides have the same sign
2 Where one side is positive and the other equation is negative

1)
4x-3y+2=3x+4y-7
x-7y=-9

2)

-(4x-3y+2)=3x+4y-7

0=7x+y-5

By using simulatenous equations we solve

x=0.52 and y=1.36 as the two locuses.

 

InteGrand

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1 Where both sides have the same sign
2 Where one side is positive and the other equation is negative

1)
4x-3y+2=3x+4y-7
x-7y=-9

2)

-(4x-3y+2)=3x+4y-7

0=7x+y-5

By using simulatenous equations we solve

x=0.52 and y=1.36 as the two locuses.

 

davidgoes4wce

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OK thanks so I did a bit too much working out , the correct locuses are as follows (from the graph):














It's the first time I have done a locus question that involved 2 equations in their solution.
 

Green Yoda

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Q2: My attempt
m(PA)=2m(PB)
(y+2)/(x-3)=2((y-7)/(x+1))
(y+2)/(x-3)=(2y-14)/(x+1)
(y+2)(x+1)=(2y-14)(x-3)
 
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