Point and a line

The original line was 6x- y= -3 which is the same as y= 6x+ 3. That has slope 6 so a perpendicular line has slope -1/6. We want that perpendicular line to go through the point (6, 2) so the perpendicular line is given by y= -(1/6)(x- 6)+ 2= -(1/6)x+ 3. Subtracting the equations of the line eliminates y giving 0= (37/6)x. At the point of intersection x= 0, y= 3. Yes, the foot of the perpendicular is (0, 3). The distance between (0, 3) and (6, 2) is indeed \(\displaystyle \sqrt{(6- 0)^2+ (2- 3)^2}= \sqrt{37}\). You have correctly solved this problem!
 
Harry_the_cat said:
Don't lose the bigger picture.
You are finding the point of intersection of the two lines y=6x+3 into x+6y=18.
By solving the equations simultaneously (by substitution) you have found that x=0 and y=3.
So what is the point of intersection?
Click to expand...
(0,3) and (6,2)?
 
The shortest distance between a point and a line (which is what is meant by "the" distance between them) is along a line perpendicular to the line. The line is given by y= 6x+ 3 so has slope 6. A line perpendicular to that has slope -1/6. The line through (6, 2) with slope -1/6 is y= (-1/6)(x- 6)+ 2 so 6y= -x+ 6+ 12 or x+ 6y= 18

Substitute (two lines which intersect) y=6x+3 into x+6y=18 and you get:
x + 6(6x+3) = 18.
x + 36x +18 = 18
37x +18 =18
37x = 0
x = 0
So that means that x=0 at the point where the lines y=6x+3 into x+6y=18 intersect.
Substitute y = 6x + 3
y = 6(0) + 3 = 3
y = 3
The point of intersection is x = 0, y = 3.

Then, use the distance formula to find the exact distance between (0,3) and 6,2)
d^2 = (x2 - x1)^2 + (y2 - y1)^2
The foot of the perpendicular to the line Is (0, 3). So the distance between (0, 3) and (6, 2), squared, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)
 
rachelmaddie said:
The shortest distance between a point and a line (which is what is meant by "the" distance between them) is along a line perpendicular to the line. The line is given by y= 6x+ 3 so has slope 6. A line perpendicular to that has slope -1/6. The line through (6, 2) with slope -1/6 is y= (-1/6)(x- 6)+ 2 so 6y= -x+ 6+ 12 or x+ 6y= 18

Substitute (two lines which intersect) y=6x+3 into x+6y=18 and you get:
x + 6(6x+3) = 18.
x + 36x +18 = 18
37x +18 =18
37x = 0
x = 0
So that means that x=0 at the point where the lines y=6x+3 into x+6y=18 intersect.
Substitute y = 6x + 3
y = 6(0) + 3 = 3
y = 3
The point of intersection is x = 0, y = 3. The point of int is (0, 3).

Then, use the distance formula to find the exact distance between (0,3) and 6,2)
d^2 = (x2 - x1)^2 + (y2 - y1)^2
The foot of the perpendicular to the line Is (0, 3). previous sentence is unnecessay So the distance between (0, 3) and (6, 2), squared, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37)
Click to expand...
Yes good. See comments in red. Make sure you understand everything you have written.
 
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