Question

The distance from a line to (6,2) is 3 units. The line is perpendicular to 4x+3y+8 =0. Find its equation.

please.. i really need your help. :)
thank you. :)

Answer 1

found it

Answer 2

i would start by perping the given line to see what its parallel to

Answer 3

4x + 3y = -8

3x - 4y = -8 is perped.

Answer 4

i have an idea from calculus ... take a circle centered at 6,2 of radius 3 and find the point such that the slope of the tangent line is 3/4

Answer 5

that dint work out too well. might as well go with the geometry setup

Answer 6

ok.:) and my other classmate said that she solved for the negative reciprocal slope of the line and i dont know if its correct.

Answer 7

4x+3y+8=0

4(6) + 3(2) = 24 + 6 = 30

4x + 3y = 30 is the line thru 6,2 lets put a circle at 6,2 now to develop a point

let y=(30-4x)/3

(x-6)^2 + (y-2)^2 = 9

(x-6)^2 + ((30-4x)/3-2)^2 = 9

(x^2 -12x +36) + (4(6-x)/3)^2 = 9

(x^2 -12x +36) + 16(x^2-12x+36)/9 = 9

9(x^2 -12x +36) + 16(x^2-12x+36) = 81

lets just say this eventually works out to x=21/5 or 39/5
y= 22/5 or -2/5

seems like 21/5, 22/5 have it. soo


3x-4y=-5 seems to fit if i didnt mess it up lol

Answer 8

what solution did yall come up with?