What's the distance between (-6,1) and y=6-3x

Answer 1

First, find an equation for a line perpendicular to y = 6 - 3x. This line has the form y = x/3 + b.

Next, we want to find the perpendicular line that passes through (-6,1).

1 = -6/3 + b so that b = 3. Thus, this line is y = x/3 + 3.

Now, find where these two lines intersect. 6 - 3x = x/3 + 3, so that the x-coordinate of intersection is x = 9/10. The y-coordinate of intersection is 3/10 + 3 = 33/10.

Finally the distance between th point and the line is the distance between the points (-6,1) and (9/10,33/10).

Answer 2

We need to find the equation of a line

perpendicular to the line

that goes through the point.

Slope of perpendicular line is 1/3

So the equation is: y = 1/3 + b

To find the "b" use (x,y) values of given point.

1 = (1/3)(-6) + b

1 = -2 + b

3 = b

So, the equation of the perpendicular line

through the point is: y = (1/3)x + 3

Now, find the point, where the perpendicular line

meets the given line:

y = y

(1/3)x + 3 = 6 - 3x

(10/3)x = 3

x = 9/10

When: x = 9/10

Then: y = 6 - 3(9/10)

y = 6 - 27/10

y = 60/10 - 27/10 = 33/10

The perpendicular line,

through the given point,

intersects the given line at point: (9/10, 33/10)

Now, we just need to find the distance

between two points: (-6, 1) and (9/10, 33/10)

D = sqrt [ (deltaX)^2 + (deltaY)^2 ]

D = sqrt [ (9/10 + 6)^2 + (33/10 - 1)^2 ]

D = sqrt [ (69/10)^2 + (23/10)^2 ]

D = sqrt [ 5290/100 ]

D = sqrt(5290) / 10