Find the distance AB between points A and B (0,1) when lines y=x+4 and 3x+2y=13 intersect at point A

Answer 1

First, we need to find the coordinates of point A where the two lines intersect.

We have the equations:
y = x + 4
3x + 2y = 13

Substitute y in the second equation with x + 4:
3x + 2(x + 4) = 13
3x + 2x + 8 = 13
5x + 8 = 13
5x = 5
x = 1

Now that we have x = 1, we can find y by substituting it back into the first equation:
y = 1 + 4
y = 5

So point A is (1, 5).

To find the distance AB, we can use the distance formula:

Distance AB = √((x2 - x1)^2 + (y2 - y1)^2)
= √((0 - 1)^2 + (1 - 5)^2)
= √((-1)^2 + (-4)^2)
= √(1 + 16)
= √17

Therefore, the distance AB between points A and B is √17 or approximately 4.12 units.