Find the equation of the lines through (7,-4) passing at a distance 1 from the point (2,1).I got this problem from the book Analytic geometry by...

We'll write the formula that represents the distance from
a point to a line.

ax + by + c = 0 and the point
(2,1)

d = |2a + b + c|/sqrt(a^2 +
b^2)

But the distance is of
1.

1 = |2a + b + c|/sqrt(a^2 +
b^2)

sqrt(a^2 + b^2) = |2a + b +
c|

We also know that the line is passing through
(7,-4):

7a - 4b + c = 0

c = 4b
- 7a

2a + b + c = 2a + b + 4b -
7a

2a + b + c = -5a +
5b

sqrt(a^2 + b^2) = -5a +
5b

We'll raise to square both
sides:

a^2 + b^2 = 25(b -
a)^2

a^2 + b^2 = 25b^2 - 50ab +
25a^2

24(a^2 + b^2) = 50
ab

12(a^2 + b^2) = 25ab

If a =
3 and b = 4

12*(3^2 + 4^2) =
25*3*4

12*(9+16) = 25*12

12*25
= 25*12

Since the multiplication is commutative, then the
identity is true for a = 3 and b = 4.

The
equation of the line 3x + 4y - 5 = 0, that is passing through the point (7 , -4) and it
is at the distance of 1 from the point (2,1) has to respect the condition between
coefficients 12(a^2 + b^2) = 25ab.