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Home > Geometry calculators > Coordinate Geometry > Find the point on the y-axis which is equidistant from A(-2,3) and B(5,4) calculator
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Solution
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Solution provided by AtoZmath.com
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point on the y-axis which is equidistant from A(-2,3) and B(5,4) calculator
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1. Find the point on the y-axis which is equidistant from `A(6,5)` and `B(-4,3)`
2. Find the point on the y-axis which is equidistant from `A(5,-2)` and `B(-3,2)`
3. Find the point on the y-axis which is equidistant from `A(-4,3)` and `B(5,2)`
4. Find the point on the y-axis which is equidistant from `A(6,-1)` and `B(2,3)`
5. Find the point on the y-axis which is equidistant from `A(2,3)` and `B(-4,1)`
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Example1. Find the point on the y-axis which is equidistant from `A(6,5)` and `B(-4,3)`Solution:Let `P(0,y)` be the point on Y-axis, which is the equidistance form `A(6,5)` and `B(-4,3)`. `:. PA=PB` `=>PA^2=PB^2` `=>(0-6)^2+(y-5)^2=(0+4)^2+(y-3)^2` `=>y^2-10y+25+36=y^2-6y+9+16` `=>-10y+6y=9+16-25-36` `=>-4y=-36` `=>y=9` Thus, the required point is `P(0,9).`
2. Find the point on the y-axis which is equidistant from `A(5,-2)` and `B(-3,2)`Solution:Let `P(0,y)` be the point on Y-axis, which is the equidistance form `A(5,-2)` and `B(-3,2)`. `:. PA=PB` `=>PA^2=PB^2` `=>(0-5)^2+(y+2)^2=(0+3)^2+(y-2)^2` `=>y^2+4y+4+25=y^2-4y+4+9` `=>4y+4y=4+9-4-25` `=>8y=-16` `=>y=-2` Thus, the required point is `P(0,-2).`
3. Find the point on the y-axis which is equidistant from `A(-4,3)` and `B(5,2)`Solution:Let `P(0,y)` be the point on Y-axis, which is the equidistance form `A(-4,3)` and `B(5,2)`. `:. PA=PB` `=>PA^2=PB^2` `=>(0+4)^2+(y-3)^2=(0-5)^2+(y-2)^2` `=>y^2-6y+9+16=y^2-4y+4+25` `=>-6y+4y=4+25-9-16` `=>-2y=4` `=>y=-2` Thus, the required point is `P(0,-2).`
4. Find the point on the y-axis which is equidistant from `A(6,-1)` and `B(2,3)`Solution:Let `P(0,y)` be the point on Y-axis, which is the equidistance form `A(6,-1)` and `B(2,3)`. `:. PA=PB` `=>PA^2=PB^2` `=>(0-6)^2+(y+1)^2=(0-2)^2+(y-3)^2` `=>y^2+2y+1+36=y^2-6y+9+4` `=>2y+6y=9+4-1-36` `=>8y=-24` `=>y=-3` Thus, the required point is `P(0,-3).`
5. Find the point on the y-axis which is equidistant from `A(2,3)` and `B(-4,1)`Solution:Let `P(0,y)` be the point on Y-axis, which is the equidistance form `A(2,3)` and `B(-4,1)`. `:. PA=PB` `=>PA^2=PB^2` `=>(0-2)^2+(y-3)^2=(0+4)^2+(y-1)^2` `=>y^2-6y+9+4=y^2-2y+1+16` `=>-6y+2y=1+16-9-4` `=>-4y=4` `=>y=-1` Thus, the required point is `P(0,-1).`
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