# What is the distance between the point (0;1) and the line that passes through the points (1;2) and (-1;-2) ?

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### 2 Answers

To find the distance from (0,1) to the line joining (1,2) and (-1,-2).

We know that the line through any two points (x1, y1) and (x2, y2) is:

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

Therefore the line through the two given points (1,2) and (-1 , -2) is:

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

y-2 = 2(x-1)

0 = 2x-y -2+2.

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

The distance from a point ( x1, y1 ) to a line ax+by +c = 0 is given by:

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

Therefore distance from (0, 1) to the line 2x-y = 0 is

d = |2*0-1|/sqrt(2^2+(-1)^2) = 2/sqrt(5) = (2sqrt5)/5

Therefore the distance from (0,1) to the line 2x-y = 0 is (2sqrt5)/5.

To calculate the distance from the point A(0;1) to the line that 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 = 2

x2 = -1, y2 = -2

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

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

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

We'll divide by -2:

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

We'll cross multiply:

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

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

2x-2 = y-2

We'll subtract y-2 both sides:

2x - 2 - y + 2 = 0

We'll combine and eliminate like terms:

2x - y = 0

The equation of the line is:

2x - y = 0

Now, we'll write the formula for distance:

d = |2*xA - 1*yA + 0|/sqrt(2^2 + (-1)^2)

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

d = 1/sqrt5

**d = sqrt5/5**