find the distance from the line 4x+3y=5 to the point (6,9)
distance equation?
m=-4/3 y-9=-4/3(x-6) y-9=-4/3x+8 y=-4/3x +17 (0, 5/3) and (6,9) d=s.rt (6-0)^2+(9-5/3)^2 =9.47511
perhaps try this.
The distance from a point P(x,o,yo) to a line ax+by+c = 0 is \[d=\frac{ \left| ax _{o}+by _{o}+c \right| }{ \sqrt{a^2+b^2} }\]
\[d=\frac{ \left| 4(6)+3(9)-5 \right| }{ \sqrt{4^2+3^2} }\]
been a while since I've done this so please correct me if i'm wrong
i got 9.2... but the hint say first find the line through the point (6,9) parallel to the given line, and then find the distance between these lines
you made a slight error on the slope of the line perpendicular to the given line its slope is - 1 / (-4/3) = 3/4
- oh the parallel line ! sorry
so was it right or wrong?
chrs00's method is correct and that comes to 9.2
yep
why did you use (0,5/3) on the original line?
i think he just used any point on that line. i think you need to use a point that is perpendicular to that line
distance=shortest distance
are u sure ..this question is structure different frm the previous set of questions that i did..they want .. to find distance frm the point (a,b) eg.(4,8) to the line y=sumthing eg 6x+8y=9 ... inwhich i used that formula provide by chrs00 but the question ask to find the distance frm the line given to the point given...
Chris00 formula gives you just that
you were given a point that did not lie on that line
another way to do it is to find the equation of the line perpendicular to the original line and passing through the point (6,9) then solving the 2 equations simultaneously to find the point of intersection. Then use the distance formula with this point and (6,9) But that's a bit long-winded.
ok
can u sketch it?
you can sketch it on the desmos web site
but i wouldnt get desmos to use in my test
the 2 parallel lines will look something like |dw:1444387746919:dw|
ok
the line joining the point (0, 5/3) and (6,9) is not perpendicular to the 2 parallel lines. That is why you didn't get the right answer
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