3. Find the point on the x-axis which is equidistant from A(5,4) and B(-2,3)
1. Find the point on the x-axis which is equidistant from `A(5,4)` and `B(-2,3)`
Solution:
Let `P(x,0)` be the point on X-axis, which is the equidistance form `A(5,4)` and `B(-2,3)`.
`:. PA=PB`
`=>PA^2=PB^2`
`=>(x-5)^2+(0-4)^2=(x+2)^2+(0-3)^2`
`=>x^2-10x+25+16=x^2 +4x+4+9`
`=>-10x-4x=4+9-25-16`
`=>-14x=-28`
`=>x=2`
Thus, the required point is `P(2,0)`.
2. Find the point on the x-axis which is equidistant from `A(2,-5)` and `B(-2,9)`
Solution:
Let `P(x,0)` be the point on X-axis, which is the equidistance form `A(2,-5)` and `B(-2,9)`.
`:. PA=PB`
`=>PA^2=PB^2`
`=>(x-2)^2+(0+5)^2=(x+2)^2+(0-9)^2`
`=>x^2-4x+4+25=x^2 +4x+4+81`
`=>-4x-4x=4+81-4-25`
`=>-8x=56`
`=>x=-7`
Thus, the required point is `P(-7,0)`.
3. Find the point on the x-axis which is equidistant from `A(7,6)` and `B(-3,4)`
Solution:
Let `P(x,0)` be the point on X-axis, which is the equidistance form `A(7,6)` and `B(-3,4)`.
`:. PA=PB`
`=>PA^2=PB^2`
`=>(x-7)^2+(0-6)^2=(x+3)^2+(0-4)^2`
`=>x^2-14x+49+36=x^2 +6x+9+16`
`=>-14x-6x=9+16-49-36`
`=>-20x=-60`
`=>x=3`
Thus, the required point is `P(3,0)`.
4. Find the point on the x-axis which is equidistant from `A(-2,5)` and `B(2,-3)`
Solution:
Let `P(x,0)` be the point on X-axis, which is the equidistance form `A(-2,5)` and `B(2,-3)`.
`:. PA=PB`
`=>PA^2=PB^2`
`=>(x+2)^2+(0-5)^2=(x-2)^2+(0+3)^2`
`=>x^2+4x+4+25=x^2 -4x+4+9`
`=>4x+4x=4+9-4-25`
`=>8x=-16`
`=>x=-2`
Thus, the required point is `P(-2,0)`.
5. Find the point on the x-axis which is equidistant from `A(5,-2)` and `B(-3,2)`
Solution:
Let `P(x,0)` be the point on X-axis, which is the equidistance form `A(5,-2)` and `B(-3,2)`.
`:. PA=PB`
`=>PA^2=PB^2`
`=>(x-5)^2+(0+2)^2=(x+3)^2+(0-2)^2`
`=>x^2-10x+25+4=x^2 +6x+9+4`
`=>-10x-6x=9+4-25-4`
`=>-16x=-16`
`=>x=1`
Thus, the required point is `P(1,0)`.
This material is intended as a summary. Use your textbook for detail explanation.
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