HW HELP : Distance Question?

Question

Find the distance from point (3,0) to the line y = -x

Answer 1

The perpendicular distance, I presume. The equation of the line perpendicular to y = -x has a slope of 1, and an equation of

y = x + b

where b remains to be found. Since the perpendicular goes through (3, 0), b = -3

We now have to find the point on the line y = -x that is nearest to (3, 0). We do this by finding where y = -x intersects with its perpendicular. We have

y = -x
y = x - 3

Thus,

-x = x - 3
2x = 3
x = 3/2, y = -3/2

Now we have to find the distance between (3, 0)
and (3/2, -3/2). You can see that the lines and points we have been working with form a 45-45-90 triangle with a hypotenuse of 3, and the distance we are looking for is one leg, or 3/sqrt(2), better written as 3sqrt(2)/2. yljacktt's answer, when simplified, is the same as this.

Minii's answer is wrong. You can see that her formula contains sqrt(2) in the denominator, but her answer does not.

Answer 2

Here is one way to solve it.
First, the slope of y=-x is m=-1, because putting it in y=mx+b form, y=(-1*x)+0.

So, the shortest distance from (3,0) to y=-x would be a line perpendicular to y=-x. So, since slope is -1, the perpendicualr slope is -(1/-1) = 1.
SO, putting in y=mx+b form, 0=(1*3)+b, so b=-3.
SO, the equation is y=x-3.

Now find point where they intersect.
x-3=-x, or 2x=3, so x=3/2. And y=-3/2, since y=-x.

SO, use distance formula d=squareroot((3-(3/2))^2+ (0--(3/2))^2) = sqr((3/2)^2+(3/2)^2)) = sqr(18/4) = 2.1213 approximately.
If I didnt make any mistakes!

Answer 3

according to formula
xa+yb+c/{(a^2+b^2)}^1/2
where (x,y) is given point a=coeff. of x of a line ,b=coeff of y
c=constant in line
acc to ques a=1,b=1,c=0,x=3,y=0
d=3(ans)

Answer 4

Plug in the number for x:
-3