The distance between a point and a line is defined to be the length of the perpendicular line segment connecting the point to the given line.
Let (x_{1},y_{1}) be the point not on the line and let (x_{2},y_{2}) be the point on the line.
To find the distance between the point (x_{1},y_{1}) and the line with equation ax + bx + c = 0, you can use the formula below.
Example #1
Find the distance between a point and a line using the point (5,1) and the line y = 3x + 2.
Rewrite y = 3x + 2 as ax + by + c = 0
Using y = 3x + 2, subtract y from both sides.
y - y = 3x - y + 2
0 = 3x - y + 2
3x - y + 2 = 0
a = 3, b = -1, and c = 2
x_{1} = 5 and y_{1} = 1
Example #2
Find the distance between a point and a line using the point (-4,2) and the line y = -2x + 5.
Rewrite y = -2x + 5 as ax + by + c = 0
Using y = -2x + 5, subtract y from both sides.
y - y = -2x - y + 5
0 = -2x - y + 5
-2x - y + 5 = 0
a = -2, b = -1, and c = 5
x_{1} = -4 and y_{1} = 2
We can redo example #1 using the distance formula. To use the distance formula, we need two points. We already have (5,1) that is not located on the line y = 3x + 2.
We can just look for the point of intersection between y = 3x + 2 and the line that is perpendicular to y = 3x + 2 and passing through (5, 1)
The line that is perpendicular to y = 3x + 2 is given by y = (-1/3)x + b.
Use the point (5, 1) to find b by letting x = 5 and y = 1.
1 = (-1/3) × 5 + b
1 = -5/3 + b
b = 1 + 5/3 = 8/3
y = (-1/3)x + 8/3
Now, set the two equations equal to themselves
(-1/3)x + 8/3 = 3x + 2
3x + (1/3)x = 8/3 - 2
(10/3)x = (8 - 6)/3
(10/3)x = 2/3
x = (2/3) × 3/10 = 2/10 = 1/5
y = 3x + 2 = 3 × 1/5 + 2 = 3/5 + 2 = 13/5
The point of intersection between y = 3x + 2 and y = (-1/3)x + 8/3 is
(1/5, 13/5) or (0.25, 2.6)
Find the distance using the points (5,1) and (0.25, 2.6).
To derive the formula at the beginning of the lesson that helps us to find the distance between a point and a line, we can use the distance formula and follow a procedure similar to the one we followed in the last section when the answer for d was 5.01.
Hang in there tight. Deriving the distance between a point and a line is among of the toughest things you have ever done in life.
The steps to take to find the formula are outlined below.
1) Write the equation ax + by + c = 0 in slope-intercept form.
2) Use (x_{1}, y_{1}) to find the equation that is perpendicular to ax + by + c = 0
3) Set the two equations equal to each other to find expressions for the points of intersection (x_{2}, y_{2})
4) Use the distance formula, (x_{1}, y_{1}), and the expressions found in step 3 for (x_{2}, y_{2}) to derive the formula.
1)
ax + by + c = 0
ax - ax + by + c = -ax
by + c = -ax
by + c - c = -ax - c
by = -ax - c
y = -ax/b - c/b
y = (-a/b)x - c/b
2) The line that is perpendicular to y = (-a/b)x - c/b can be written as
y = (b/a)x + y-intercept
Use (x_{1}, y_{1}) to find y-intercept
y_{1} = (b/a)x_{1} + y-intercept
y-intercept = y_{1}- (b/a)x_{1}
y = (b/a)x + y_{1}- (b/a)x_{1}
3) Set the two equations equal to each other to find expressions for the points of intersection (x_{2}, y_{2})
Set y = (-a/b)x - c/b and y = (b/a)x + y_{1}- (b/a)x_{1} equal to each other
(-a/b)x - c/b = (b/a)x + y_{1}- (b/a)x_{1}
(b/a)x + (a/b)x = -c/b + (b/a)x_{1} - y_{1}
(b/a)x + (a/b)x = (ba/ba) × [(-c/b + (b/a)x_{1} - y_{1}]
(b/a + a/b)x = (-ca + b^{2}x_{1} - y_{1}ba) / ba
[ (a^{2} + b^{2})/ab ] / x = (-ca + b^{2}x_{1} - y_{1}ba) / ba
x = (-ca + b^{2}x_{1} - y_{1}ba) / a^{2} + b^{2 }( this is x_{2 })
Now, let us find y_{2} using the equation y = (-a/b)x - c/b
(-a/b)x = -a/b[ (-ca + b^{2}x_{1} - bay_{1}) / (a^{2} + b^{2}) ]
(-a/b)x = (ca^{2} - ab^{2}x_{1} + ba^{2}y_{1}) / b(a^{2} + b^{2})
(-a/b)x - c/b = [(ca^{2} - ab^{2}x_{1} + ba^{2}y_{1}) / b(a^{2} + b^{2})] - c/b
(-a/b)x - c/b = (ca^{2} - ab^{2}x_{1} + ba^{2}y_{1} - ca^{2} - b^{2}c) / b(a^{2} + b^{2})
(-a/b)x - c/b = (- ab^{2}x_{1} + ba^{2}y_{1} - b^{2}c) / b(a^{2} + b^{2})
(-a/b)x - c/b = b[(- abx_{1} + a^{2}y_{1} - bc)] / b(a^{2} + b^{2})
(-a/b)x - c/b = (- abx_{1} + a^{2}y_{1} - bc) / (a^{2} + b^{2})
y = (- abx_{1} + a^{2}y_{1} - bc) / (a^{2} + b^{2}) (this is y_{2})
To summarize,
x_{2} = (-ca + b^{2}x_{1} - y_{1}ba) / a^{2} + b^{2}
y_{2} = (- abx_{1} + a^{2}y_{1} - bc) / (a^{2} + b^{2})
Now, find the distance between a point and a line using (x_{1},y_{1}) and (x_{2},y_{2})
Nov 18, 20 01:20 PM
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