The points A=(1,1) B=(3,4) connects to point C (3,1).
a. Find the length of AB
b. The gradient and equation of line AB
Given: A(1,1) B(3,4), C(3,1).
a. (AB)^2 = (3-1)^2 + (4-1)^2 = 13
AB = sqrt13.
b. m = (4-1) / (3-1) = 3/2.
Y = mx + b
1 = (3/2)*1 + b
b = 1-3/2 = -1/2.
Eq: Y = (3/2)x - 1/2.
Henry
To find the length of AB, we can use the distance formula. The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's plug in the coordinates of point A (x1, y1) = (1, 1) and point B (x2, y2) = (3, 4) into the formula:
d = sqrt((3 - 1)^2 + (4 - 1)^2)
= sqrt(2^2 + 3^2)
= sqrt(4 + 9)
= sqrt(13)
So, the length of AB is sqrt(13).
To find the gradient and equation of line AB, we can use the slope-intercept form of a linear equation, which is given by y = mx + b, where m is the gradient and b is the y-intercept.
First, let's calculate the gradient (m) of line AB. The gradient is defined as the change in y divided by the change in x between two points on the line. We'll use the coordinates of point A (x1, y1) = (1, 1) and point B (x2, y2) = (3, 4).
m = (y2 - y1) / (x2 - x1)
= (4 - 1) / (3 - 1)
= 3 / 2
= 1.5
So, the gradient of line AB is 1.5.
Next, let's find the equation of line AB using the point-slope form. We'll choose point A (1, 1) to substitute into the equation.
y = mx + b
1 = 1.5 * 1 + b
1 = 1.5 + b
b = 1 - 1.5
b = -0.5
Therefore, the equation of line AB is y = 1.5x - 0.5.