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.
1 Answer | Add Yours
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.
We’ve answered 319,864 questions. We can answer yours, too.Ask a question