Question

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

Answer 1

Answer: 28x+42y=3

Step-by-step explanation:

Given: The point (x,y) is equidistant from the points  $(\frac{-1}{4},-4)\ and\ (\frac{13}{4},\frac{5}{4})$.

By distance formula the distance between $(\frac{-1}{4},-4)\ and\ (x,y)$ is

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

Similarly, the distance between $(\frac{13}{4},\frac{5}{4})\ and\ (x,y)$ is

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

Since,

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