<u>Answer-</u>
The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are
![12x-5y+14=0 \\ 12x-5y-14=0](https://tex.z-dn.net/?f=12x-5y%2B14%3D0%20%5C%5C%2012x-5y-14%3D0)
<u>Solution-</u>
Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)
The applying the distance formula,
![\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cleft%20%7C%20%5Cfrac%7B12h-5k-1%7D%7B%5Csqrt%7B12%5E2%2B5%5E2%7D%7D%20%5Cright%20%7C%3D1)
![\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cleft%20%7C%20%5Cfrac%7B12h-5k-1%7D%7B%5Csqrt%7B169%7D%7D%20%5Cright%20%7C%3D1)
![\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cleft%20%7C%20%5Cfrac%7B12h-5k-1%7D%7B13%7D%20%5Cright%20%7C%3D1)
![\Rightarrow 12h-5k-1=\pm 13](https://tex.z-dn.net/?f=%5CRightarrow%2012h-5k-1%3D%5Cpm%2013)
![\Rightarrow 12h-5k=\pm 14](https://tex.z-dn.net/?f=%5CRightarrow%2012h-5k%3D%5Cpm%2014)
![\Rightarrow 12h-5k=14,\ 12h-5k=-14](https://tex.z-dn.net/?f=%5CRightarrow%2012h-5k%3D14%2C%5C%2012h-5k%3D-14)
![\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0](https://tex.z-dn.net/?f=%5CRightarrow%2012h-5k-14%3D0%2C%5C%2012h-5k%2B14%3D0)
![\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0](https://tex.z-dn.net/?f=%5CRightarrow%2012x-5y-14%3D0%2C%5C%2012x-5y%2B14%3D0)
Two equations are formed because one will be upper from the the given line and other will be below it.
Answer:
Step-by-step explanation:
1,8
5,7
2,3
Answer:
1st one I hope this helped!
Step-by-step explanation:
9514 1404 393
Answer:
(a) f(x) = √x -2
Step-by-step explanation:
The square root function has no horizontal shift, so the expression under the radical is simply x with no added terms. There is only one such answer choice, which also shows the necessary translation downward by an added negative constant.
![f(x)=\sqrt{x}-2](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%7Bx%7D-2)