Answer: 28x+42y=3
Step-by-step explanation:
Given: The point (x,y) is equidistant from the points [tex](\frac{-1}{4},-4)\ and\ (\frac{13}{4},\frac{5}{4})[/tex].
By distance formula the distance between [tex](\frac{-1}{4},-4)\ and\ (x,y)[/tex] is
[tex]D_1=\sqrt{(x-\frac{-1}{4})^2+(y-(-4))^2} \\=\sqrt{(x+\frac{1}{4})^2+(y+4)^2}[/tex]
Similarly, the distance between [tex](\frac{13}{4},\frac{5}{4})\ and\ (x,y)[/tex] is
[tex]D_2=\sqrt{(x-\frac{13}{4})^2+(y-(\frac{5}{4}))} \\=\sqrt{(x-\frac{13 }{4})^2+(y-\frac{5}{4})^2}[/tex]
Since,
[tex]D_1=D_2\\\\\Rightarrow\ \sqrt{(x+\frac{1}{4})^2+(y+4)^2}=\sqrt{(x-\frac{13}{4})^2+(y-\frac{5}{4})^2}\\\\\text{Squaring on the sides, we get}\\\Rightarrow(x+\frac{1}{4})^2+(y+4)^2=(x-\frac{13}{4})^2+(y-\frac{5}{4})^2\\\\\Rightarrow[x^2+\frac{1}{2}x+\frac{1}{16}]+y^2+8y+16=x^2-\frac{13}{2}x+\frac{169}{16}+y^2+\frac{25}{4}-\frac{5}{2}y\\\\\Rightarrow7x+\frac{21}{2}y=\frac{3}{4}\\\Rightarrow28x+42y=3[/tex]
