## DLS Group Title 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? one year ago one year ago

1. hartnn Group Title

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

2. DLS Group Title

How can we solve this using parametric form?

3. hartnn Group Title

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

4. hartnn Group Title

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

5. DLS Group Title

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

6. DLS Group Title

r is the distance

7. hartnn Group Title

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

8. DLS Group Title

oh sorry its 2,-1

9. DLS Group Title

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

10. DLS Group Title

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

11. DLS Group Title

indirectly

12. hartnn Group Title

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

13. hartnn Group Title

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

14. hartnn Group Title

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

15. hartnn Group Title

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

16. hartnn Group Title

its like normalization....

17. DLS Group Title

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

18. DLS Group Title

any special logic or anything?

19. hartnn Group Title

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(....)

20. hartnn Group Title

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

21. DLS Group Title

seems legit..thanks :D

22. hartnn Group Title

welcome ^_^