Question

Given the points A(-2, -1) and B(-5, 3), answer the following questions.

a) Find the length of the segment.

b) Find the slope of the line segment.

c) Write the equation of the line that contains the two points.

Answer 1

a) 5

b) -4/3

c) y = -4/3x - 11/3

Explanation:

The length of the segment means the length in between two points. Use 2 dimensional distance formula.

`d = \sqrt{(x_(2)-x_(1))^2 + (y_(2)-y_(1))^2 }`

Plug in your points.

x value is the first point in the set, the horizontal measure in units.

y value is the second point in the set, the vertical measure in units.

x values of 2 points: -2, -5

y values of 2 points: -1, 3

`d = √((-5--2)^2 + (3--1)^2 )`

A negative value subtracted by another negative value is the first negative value plus the inverse of the second negative value.

For example,

-5 - -6 = -5 + 6 = 1

Simplify.

`d = √((-3)^2 + (4)^2 )`

`d = √(9 + 16)`

`d = √(25)`

5

Square root of 25 is 5 because 5^2, which is 5 x 5 is 25.

The slope is how much the y-value changes when the x-value increases by 1.

To calculate this, plug your points into the slope formula.

`Slope = (y_2-y_1)/(x_2-x_1)`

`(3--1)/(-5--2)`

`(4)/(-3)`

-4/3

I'm going to assume that the problem is asking you to write it in the form of y=mx+b, or slope-intercept form, a way of easily graphing a line. If it is asking you to express it in another way like Standard or Point-Slope Form, comment and I'll change it.

y = mx+b

m = slope

b = y-intercept, or the y-value of when the x-value is 0 on a line

y and x stays as variables if you want to graph it and for generally answering problems.

So, if x = 0, we can plug it into what we currently know.

We know the points: (-2,-1) or (-5,3)

Plug in either set of points into the slope-intercept equation.

I'll use -2,-1 but you can use either.

-1 = (-4/3) (-2) + b

Because we're solving for when x is 0, replace -2 for 0.

-1 = 8/3 + b

Subtract 8/3 on both sides.

-1 - 8/3 = 8/3 - 8/3 + b

-1 - 8/3 = b

b = -11/3

Now, the equation is:

y = -4/3x - 11/3

This is the beauty of the forms of linear equations. You can plug in any points to solve for the full set of points. You can convert any form of the linear equation to another form.