The distance from a point to a line is defined as the shortest distance. This is going to be the segment that runs perpendicular to the line.
Lines that run to each other have slopes that are opposite reciprocals. This means the product of their slopes will equal -1.
m_1* m_2=- 1
Examining the slope-intercept form of the line, we know that it has a of - ab. By substituting this value into the formula, we can find the slope of the perpendicular line.
The slope of the perpendicular line is ba. To find the equation of the line, we substitute the known point in the slope-intercept form, setting the slope to ba, and then solve for the y-intercept.
y=b/ax+c
y_0=b/a x_0+c
y_0-b/ax_0=c
c=y_0-b/ax_0
c=y_0-bx_0/a
c=ay_0/a-bx_0/a
c=ay_0-bx_0/a
Now we can write the equation of the perpendicular line.
y=b/ax+ay_0-bx_0/a
Finding the Point of Intersection
To find our distance, we also need to know where the given line and the perpendicular line intersect. By setting up a , we can find the point of intersection.
y=- abx & (I) y= bax+ ay_0-bx_0a & (II)
Since both equations have y isolated, it's convenient to use the . We will substitute the value of y from Equation (I) into Equation (II).
y=b/ax+ay_0-bx_0/a
- a/bx=b/ax+ay_0-bx_0/a
- a/bx-b/ax=ay_0-bx_0/a
x(- a/b-b/a)=ay_0-bx_0/a
x=(ay_0-bx_0/a)/(- a/b-b/a)
From here, we subtract the fractions in the denominator before multiplying the numerator and the denominator by ab.
x=(ay_0-bx_0/a)* ab/(- a/b-b/a) * ab
⇔
x=b(ay_0-bx_0)/- a^2-b^2
Now, we simplify a little more by multiplying the numerator and denominator by -1.
x=b(ay_0-bx_0)* (-1)/(- a^2-b^2)* (-1)
⇔
x=b(bx_0-ay_0)/a^2+b^2
Having solved the second equation for x, we can substitute this value into the first equation to find the value of y.
y=- a/bx
y=- a/b* b(bx_0-ay_0)/a^2+b^2
y=- ab(bx_0-ay_0)/b(a^2+b^2)
y=- a(bx_0-ay_0)/a^2+b^2
y=- a(bx_0-ay_0)/a^2+b^2
y=a(ay_0-bx_0)/a^2+b^2
Finding the Distance
When we know the coordinates of the two points making up the segment, we can use the to calculate it.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
d=sqrt(( b(bx_0-ay_0)/a^2+b^2- x_0)^2+( a(ay_0-bx_0)/a^2+b^2- y_0)^2)
d=sqrt((b(bx_0-ay_0)/a^2+b^2-x_0(a^2+b^2)/a^2+b^2)^2+(a(ay_0-bx_0)/a^2+b^2-y_0(a^2+b^2)/a^2+b^2)^2)
d=sqrt((b^2x_0-aby_0/a^2+b^2-a^2x_0+b^2x_0/a^2+b^2)^2+(a^2y_0-abx_0/a^2+b^2-a^2y_0+b^2y_0/a^2+b^2)^2)
d=sqrt((- aby_0-a^2x_0/a^2+b^2)^2+(- abx_0-b^2y_0/a^2+b^2)^2)
To simplify this further, there are two things we need to do. First, remember that the square of a negative is a positive number. Then, we can use the properties of perfect square trinomials.
d=sqrt((- aby_0-a^2x_0/a^2+b^2)^2+(- abx_0-b^2y_0/a^2+b^2)^2)
d=sqrt((- (aby_0+a^2x_0)/a^2+b^2)^2+(- (abx_0+b^2y_0)/a^2+b^2)^2)
d=sqrt((- aby_0+a^2x_0/a^2+b^2)^2+(-abx_0+b^2y_0/a^2+b^2)^2)
d=sqrt((aby_0+a^2x_0/a^2+b^2)^2+(abx_0+b^2y_0/a^2+b^2)^2)
d=sqrt((aby_0+a^2x_0)^2/(a^2+b^2)^2+(abx_0+b^2y_0)^2/(a^2+b^2)^2)
d=sqrt((aby_0+a^2x_0)^2+(abx_0+b^2y_0)^2/(a^2+b^2)^2)
d=sqrt(a^2b^2y_0^2 + 2a^3x_0by_0 + a^4x_0^2+a^2x_0^2b^2+2ax_0b^3y_0+b^4y_0^2/(a^2+b^2)^2)
Finally, we can remove common factors from the numerator and the denominator before fully simplifying the distance.
d=sqrt(a^2b^2y_0^2 + 2a^3x_0by_0 + a^4x_0^2+a^2x_0^2b^2+2ax_0b^3y_0+b^4y_0^2/(a^2+b^2)^2)
d=sqrt(a^2( b^2 y_0^2 + 2ax_0b y_0 + a^2 x_0^2)+a^2x_0^2b^2+2ax_0b^3y_0+b^4y_0^2/(a^2+b^2)^2)
d=sqrt(a^2( b^2 y_0^2 + 2ax_0by_0 + a^2 x_0^2)+b^2(a^2x_0^2+2ax_0by_0+b^2 y_0^2)/(a^2+b^2)^2)
d=sqrt(a^2(a^2 x_0^2 + 2ax_0by_0 + b^2 y_0^2)+b^2(a^2x_0^2+2ax_0by_0+b^2 y_0^2)/(a^2+b^2)^2)
d=sqrt((a^2+b^2)(a^2 x_0^2 + 2ax_0by_0 + b^2 y_0^2)/(a^2+b^2)^2)
d=sqrt(a^2 x_0^2 + 2ax_0by_0 + b^2 y_0^2/a^2+b^2)
d=sqrt((ax_0 + by_0)^2/a^2+b^2)
d=sqrt((ax_0 + by_0)^2)/sqrt(a^2+b^2)
d=|ax_0 + by_0|/sqrt(a^2+b^2)
Having found the formula, we can try it using the point (1,1) and the line 2x+3y=0. This means a= 2, b= 3, x_0= 1, and y_0=1
d=|ax_0 + by_0|/sqrt(a^2+b^2)
d=| 2( 1) + 3(1)|/sqrt(2^2+ 3^2)
d=|2 + 3|/sqrt(2^2+ 3^2)
d=|2 + 3|/sqrt(4+9)
d=|5|/sqrt(13)
d=5/sqrt(13)
d≈ 1.39
The distance is approximately 1.39 units.
