Find the distance AB between points A and B (5,4) when lines y=x+1 and 3x+4y=11 intersect at point A
First, find the coordinates of point A where the two lines intersect by solving the system of equations:
y = x + 1
3x + 4y = 11
Substitute the first equation into the second equation:
3x + 4(x + 1) = 11
3x + 4x + 4 = 11
7x + 4 = 11
7x = 7
x = 1
Now, substitute x = 1 into the first equation to find y:
y = 1 + 1
y = 2
So, point A is (1, 2).
Now, calculate the distance AB using the distance formula:
AB = √[(x2 - x1)^2 + (y2 - y1)^2]
AB = √[(5 - 1)^2 + (4 - 2)^2]
AB = √[4^2 + 2^2]
AB = √(16 + 4)
AB = √20
AB ≈ 4.47
Therefore, the distance AB between points A and B when lines y = x + 1 and 3x + 4y = 11 intersect at point A is approximately 4.47.