Question

Can someone help me for this question? I need to how I need to solve?! Thank you

Answer 1

Answer: 28x+42y=3

Explanation:

Given: The point (x,y) is equidistant from the points
`((-1)/(4),-4)\ and\ ((13)/(4),(5)/(4))`.

By distance formula the distance between
`((-1)/(4),-4)\ and\ (x,y)` is

`D_1=\sqrt{(x-(-1)/(4))^2+(y-(-4))^2} \\=\sqrt{(x+(1)/(4))^2+(y+4)^2}`

Similarly, the distance between
`((13)/(4),(5)/(4))\ and\ (x,y)` is

`D_2=\sqrt{(x-(13)/(4))^2+(y-((5)/(4)))} \\=\sqrt{(x-(13 )/(4))^2+(y-(5)/(4))^2}`

Since,

`D_1=D_2\\\\\Rightarrow\ \sqrt{(x+(1)/(4))^2+(y+4)^2}=\sqrt{(x-(13)/(4))^2+(y-(5)/(4))^2}\\\\\text{Squaring on the sides, we get}\\\Rightarrow(x+(1)/(4))^2+(y+4)^2=(x-(13)/(4))^2+(y-(5)/(4))^2\\\\\Rightarrow[x^2+(1)/(2)x+(1)/(16)]+y^2+8y+16=x^2-(13)/(2)x+(169)/(16)+y^2+(25)/(4)-(5)/(2)y\\\\\Rightarrow7x+(21)/(2)y=(3)/(4)\\\Rightarrow28x+42y=3`