Given the points (1,2) and (3,4), calculate the distance from (0,0) to the line that passes through the points (1,2) and (3,4).

Expert Answers
hala718 eNotes educator| Certified Educator

First we need to find the equation for the line passes through (1,2) and (3,4)

y-y1= m(x-x1)

m= (y2-y1)/(x2-x1) = (4-2)/(3-1) = 2/2 = 1

Then:

y-2= 1(x-1)

y= x-1 + 2

y= x+1

y-x-1=0

Now we need to find the distance between the line y-x-1=0 and the point (0,0)

d= l(ax+by + cl/sqrt(a^2 + b^2)

  = 0+0 -1/sqrt(1+1) = l-1l/sqrt2 = 1/sqrt2= sqrt2/2

Then the distance is '2sqrt/2' units.

neela | Student

The equation of the line joining (x1, y1) and (x2,y2)  is:

y-y1 = (y2-y1)/(x2-x1) .  Given (x1,y1)  = (1,2) and (x2,y2) = (3,4).

So the equation of the line  joining (1,2) (3,4) is:

y-2 = (4-2)/((3-1) {x-1}

y-2 = (x-1)

0 = x-y-1+2

x-y+1 = 0.....................(1)

The distance d of the  the point (x1, y1) from the line ax+bx+c is given by:

d = | |(ax1+by1+c)/sqrt(a^2+b^2)|

 Therefore , the distance d of  (0 , 0 ) from the line at (1) is:

d |( 1*00-1*0+1)/sqrt[1^2+(-1)^2] = 1/sqrt2.

giorgiana1976 | Student

The distance from the origin to the line that passes through the given points is the perpendicular to that line.

We have to determine the equation of the line that passes through the given points.

Because we know the coordinates of the points, we'll calculate the equation of the line in this way:

(x-x1)*(y2-y1)=(y-y1)(x2-x1)

(x-1)(4-2)=(y-2)(3-1)

2(x-1)=2(y-2)

We'll divide by 2 both sides:

x-1 = y-2

The equation will become:

l: y-x-1=0

The distance between a point and a line is:

d(O,l)=|(-1*0+1*0-1)|/sqrt[(-1)^2+(1)^2]

d(O,l) = |-1|/sqrt 2

d(O,l) = (sqrt 2)/2

thewriter | Student

The equation of the line between (1,2) and (3,4) is

y-2={(4-2)/(3-1)}(x-1)=(x-1)

Now y-2=x-1 can be written in the form y=x+1.

The distance of a point (a,b) from a line y=mx+c is

|b-ma-c|/((m^2+1))^(1/2)

The distance of (0,0) from the line y=x+1 is:

|0-0-1|/((1^2+1))^(1/2)= 1/sqrt 2