# Thread: Help with the distance formula i have attached a photo see the question

2. ## Re: Help with the distance formula i have attached a photo see the question

As an example this is part
Rearrange line to y=4x/3 +14/3 So gradient =4/3 Hence perpendicular will have gradient -3/4
Let (a,b) be foot of the perpendicular. Then line joining (-3,-1) to (a,b) has gradient -3/4
So (b+1)/(a+3)=-3/4 This rearranges to 4b+3a=-13 But (a,b) is on the original line so 3b-4a-14=0
Solving these simultaneous equations get a=-19/5 b==-2/5
So for answer you want distance between (-3,-1) and (-19/5,-2/5)

3. ## Re: Help with the distance formula i have attached a photo see the question

1. First find the perpendicular function's formula.
y=ax+b the slope will be -1/a
y1=2x-1 becomes y2=-x/2+b

For the value of b replace y and x by the point given (0,0)
0=-0/2+b gives you b=0
y2=-x/2

2. Find the intersection of both functions (at the intersection y1=y2)
2x-1=-x/2 then 2/5=x
y=2(2/5)-1 then -1/5=y

3. With this new point (2/5,-1/5) you can find the distance from the given point (0,0) origin.

Distance d= sqrt[(x2-x1)2+(y2-y1)2]

or the pythagorean: a2=b2+c2

Let's say that b=x c=y

So (2/5)2+(-1/5)2= a2

For the other problems use the distance formula since it's not (0,0) I hope this was helpful you could use an other formula that wiki has but I hate memorizing 100 formulas so I use the basic ones.

4. ## Re: Help with the distance formula i have attached a photo see the question

The shortest distance from the point to a line is on the perpendicular to the line so define the equation of the perpendicular line and then see where these lines cross. When you have that point you can then find the distance from it to the origin. Lets start with the line 2x - y = 1. In slope intercept form it becomes y = 2x - 1. The line perpendicular to this line would then be y = -1/2 x +b Since the origin is on this line, b=0 and the equation becomes y = - 1/2 x. Now find where this line crosses the first line by substitution.
If y = -1/2 x then the original equation becomes 2x - (-1/2 x) =1 5/2 X = 1 X = 2/5 Substitute that into the equation y = 1/2 X and find y. So the two points are (0,0) and (2/5, 1/5) Substitute these values into the distance formula or just remember that you are solving for the hypotenuse of a right triangle and the legs are the difference in the x values and the y values. Therefore, (2/5)^2 + (1/5)^2 = d^2 4/25 + 1/25 = 5/25 = 1/5 = d^2 d = square root 5/5

5. ## Re: Help with the distance formula i have attached a photo see the question

Originally Posted by roger1505
Given any line $\ell:Ax+By+C=0$ where $|A|+|B|\ne 0$ and a point $P(p,q)$ then the distance from $P\text{ to }\ell$ is $\frac{|Ap+Bq+C|}{\sqrt{A^2+B^2}}$.