Determine a distance from origin of x axis to the line that passes through the points (1;1) and (2;3).

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the distance from origin to the line that passes through (1,1) and (2,3) :

The line passing through (x1,y1) and (x2,y2) is

y-y1 = {(y2-y1)/(x2-x1)}{x-x1).

So the line through the given points (1,1) and (2,3) is:

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

y -1 = 2(x-1)

2x-y -2+1 = 0

2x-y-1 = 0.................(1).

The distance d of the line ax+by+c = 0 from the origin is given by:

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

Therefore the distance of the 2x-y-1 = 0 from the origin is:

d = |-1/(sqrt(2^2+(-1)^2)| = 1/sqrt5

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To calculate the distance from the origin (0,0) to the line taht passes through the given points, we'll have to determine first the equation of the line.

The equation of the line that passes through the points is:

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

x1 = 1, y1 = 1

x2 = 2, y2 = 3

We'll substitute the values and we'll get:

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

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

We'll cross multiply:

2(x-1) = y-1

We'll remove the brackets and we'll have:

2x - 2 = y - 1

We'll subtract y - 1:

2x - 2 - y + 1 = 0

We'll combine like terms:

2x - y - 1 = 0

Now, we'll write the formula for distance:

d = |2*xO - 1*yO + 1|/sqrt(2^2 + (-1)^2)

d = |2*0 - 1*0 + 1|/sqrt(4+1)

d = 1/sqrt5

d = sqrt5/5

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