Determine the the shortest distance of point (3,1) to the right of the equation 3x + y = -2?

Question

Determine the the shortest distance of point (3,1) to the right of the equation 3x + y = -2, au dixieme pres.

How do I do this?

Answer 1

Transform the the equation into the form y=Mx+B
Y=-3x-2
create a formula for a generic line that is perpendicular to that line
y=-1/M*x+C
Y=1/3*x+C
solve this equation for C
C=Y-1/3x
substitute (3,1) into this equation (x,y)
C=1-1/3*(-3)=0
substituting C into the perpendicular equation gives
y=1/3x
substituting this into the original equation gives
3x+1/3x=-2
10/3x=-2
x=-3/5
y=-1/5
3+3/5=18/5=the x component of the distance
1+1/5=6/5 = the y component of the distance
18*18=324
6*6=36
5*5=25
324+36=360
360/25=72/5
the square root of 72/5 is about 3.79

Don't trust my arithmetic.

Answer 2

nicely, you're good. You first desire the line perpendicular to y = -3x - 2 that runs by the point (3,a million) m = a million/3 and (3,a million) y - a million = (a million/3)(x - 3) y = (a million/3)x Now, locate the position those 2 strains intersect. y = y (a million/3)x = -3x - 2 2 = (-10/3)x -3/5 = x (-3/5, -a million/5) Now, use the gap formulation :: d = ( (x2 - x1)^2 + (y2 - y1)^2 )^(a million/2) d = ( (3 + 3/5)^2 + (a million + a million/5)^2 )^(a million/2) d = ( (18/5)^2 + (6/5)^2 )^(a million/2) d = ( 360 )^(a million/2) / 5 = 3.7947 gadgets

Answer 3

Shortest distance is the perpendicular distance
= (ax + by + c) / sqrt(a^2 + b^2)

3x + y = -2
3x + y + 2 = 0
a = 3, b = 1 and c = 2

Distance = [(3)(3) + (1) + 2] / sqrt(3^2 + 1^2)
= 12 / sqrt(10) = 3.79

Answer 4

Well you know if you have a line and a point, the shortest distance between them will be a line perpindicular to the line going through the point.

So we have the slope of the line connecting them, 1/3, and we know it goes through (3,1) so the equation is y=x/3

Now we just find where our two lines intersect by substitution of y

3x+x/3=-2

A little manipulation gives us 10x/3=-2

multiply both sides by 3 and divide by 10 and you get x=-.6

We put this in our first equation to get y=-.2

the distance between two points is given by the formula d(p1,p2)=((x1-x2)^2+(y1-y2)^2)^.5, so simple substitution gives us around 3.87731


I think...