Question

Let (2,-1) be a point on line 3x+4y=2.Find the point at distance of 5 units on the line from the given point?

Answer 1

\(\large{distance}(ax+by+c=0, (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\)

Answer 2

How can we solve this using parametric form?

Answer 3

i didn't get your question actually,
(2,1) does NOT lie on that line.
can u draw a figure or something ?

Answer 4

Distance between points (x1,y1) and (x2,y2) is
\(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

Answer 5

\[\frac{x-x1}{\cos \Theta} = \frac{y-y1}{\sin \Theta} = r\]

Answer 6

r is the distance

Answer 7

but,(2,1) does NOT lie on that line.

Answer 8

oh sorry its 2,-1

Answer 9

3x+4y+6-6=2
3(x-2)=4(y+1)
\[\frac{x-2}{-4/5}-\frac{y+1}{3/5}=r=+5 or -5\]

Answer 10

why did we divide by sqrt of a^2+b^2 that is 5?
They say by doing that we get sin Theta and cos Theta

Answer 11

indirectly

Answer 12

so that the two numbers(-4/5 and 3/5) here satisfy, a^2+b^2=1
which is the property of sin and cos

Answer 13

so we can take one as -4/5 and other as 3/5

Answer 14

like here u take cos t =-4/5, sin t = 3/5

Answer 15

if you did not divide by 5,
then 4^2+3^2 does not equal 1
so u cannot take cos t=4, and sin t =3

Answer 16

its like normalization....

Answer 17

i know that but why did we divide by sqrt of a^2+b^2 only?

Answer 18

any special logic or anything?

Answer 19

for any 2 numbers
\(\large(\frac{a}{\sqrt{a^2+b^2}})^2+(\frac{b}{\sqrt{a^2+b^2}})^2=1\)
so there is a need to divide by sqrt(....)

Answer 20

what else can u divide those numbers by, to get their sum of squares =1 ?

Answer 21

seems legit..thanks :D

Answer 22

welcome ^_^