I'm not sure where to post this. So I'm posting in the General Math section.This is a simple coordinate geometry problem: We have to find the equation of line(s) passing through the point (7,17) and having a distance of 6 units from the point(1,9). Now I'm posting my approach: the equation of line passing through(7,17) and having slope m is y-17=m(x-7) or, mx-y-7m-17=0 now the line have a distance of 6 units from point(1,9). so [itex]\frac{m-9-7m+17}{\sqrt{m^2 +1}}[/itex]=[itex]\pm[/itex]6 or,(6m-8)^{2} =36(m^{2} +1) or, m=7/24 so the equation of line becomes 7x-24y+359 =0 This way we get only line.But if you draw it in a graph paper,you'll see that there should be another line which is x-7=0 which is parallel to the y-axis.(draw a circle of radius 6 from (1,9) and then draw tangent from (7,17) to the circle). I think we can't get the second one because it has undefined slope. My question is how can I get the second line without plotting in graph?
hi silent_hunter! (try using the X^{2} button just above the Reply box ) you lost half the solutions when you took the square-root of this line … you should have put the intermediate step (6m-8) = ±36(m^{2} +1)
thanks for your reply (6m-8)^{2} =36(m^{2} +1) in that step I squared both sides of the equation ,so the ± sign should go away.
sorry I made a typing mistake. Its mx-y-7m-17=0 but I didn't get what you said. Would you please elaborate?Thanks.
one of the two lines is x = 7, isn't it? (with slope ∞) that doesn't come up for any m in mx-y-7m-17=0
if you think, (y-9)^2+(x-1)^2=36 then if you think of it as (y-9)^2 + (x-1)^2 = 36 then: y=cos(3(x-1))+15.91 or so should hit twice. I'll bet some kind of absolute value function would hit three times.