If the points A(2,-4) is equidistant from P (3,8) and Q (-10,y), then find the value of y . Also , find distance PQ.

Explanation

According to the question ,
A(2,-4) is equidistant from P (3,8) =Q (-10,y) is equidistant from A (2,-4)
i.e., PA=QA
`rArrsqrt((2-3)^(2)+(-4-8)^(2))=sqrt((2+10)^(2)+(-4-y)^(2))`
[`:'` distance between two points `(x_(1),y_(1))and (x_(2),y_(2))`,d`=sqrt((x_(2)-x_(1))+(y_(2)-y_(1))^(2))]`
`rArrsqrt((-1)^(2)+(-12)^(2))=sqrt((12)^(2)+(4+y)^(2))`
`rArrsqrt(1+144)=sqrt(144+16+y^(2)+8y)`
`rArrsqrt(145)=sqrt(160+y^(2)+8y)`
On squaring both the sides , we get
`145=160+y^(2)+8y`
`rArry^(2)+8y+160-145=0`
`rArry^(2)+8y+15=0`
`rArry^(5)+5y+3y+15=0`
`rArry(y+5)+3(y+5)=0`
`rArr(y+5)(y+3)=0`
If y+5=0, then y=-5
If y+3=0, then y=-3
`:. y=-3,-5`
Now , distance between P (3,8) and Q (-10,y),
PQ `=sqrt((-10-3)^(2)+(y-8)^(2))` " " [Putting y=-3]
`=sqrt((-13)^(2)+(-3-8)^(2))`
`rArr =sqrt(169+121)=sqrt(290)`
Again , distance between P (3,8) and (-10,y),Pq `=sqrt((-13)^(2)+(-5-8)^(2))` " " [putting y=-5]
`=sqrt(169+169)=sqrt(338)`
Hence , the value of y are -3-, -5 and corresponding value of PQ are `sqrt(290)` and `sqrt(338)=13sqrt(2)` , respectively.