4. Find the point on the y-axis which is equidistant from A(6,5) and B(-4,3)
1. 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).`
This material is intended as a summary. Use your textbook for detail explanation.
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