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 love and rainville and it give me these answers:...

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 love and rainville and it give me these answers: 3x+4y=5 and 4x+3y=16. Please show me how it is solve. Please show me solutions. Hope you could help me.

Asked on by eng-bryan

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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