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 Math Help > Geometry > Points and Lines

This is a page of help for 6th, 7th, and 8th graders who need to find functions for lines that pass through certain points, or that have a certain slope, etc.

You may click one of the following problems to jump straight to it:

If you don't find what you're looking for, maybe one of the choices, above, includes what you're looking for.  For example, if you need to find the y-intercept of a line that passes through two points, you can find the function for the line, then just look at the constant term, which will be the y-intercept.  Same goes for slope, which is the x-coefficient.

### Contents of this section:

On the following pages, you may find these topics:

The Determinant Method of finding the equation of a line that goes through two points in any number of dimensions.

Graphing a line has an example of finding the slope of a line, given its equation, and then graphing it.

Lines in 3D introduces vectors and parametric representations of lines in three dimensional space.

A discussion of a Point and Triangle -- is the point inside or outside the triangle?  The challenge is to find an algorithm that answers this question.  The solution to this challenge is presented.

1. Find a function for the line containing a pair of points.
(2,6) and (6,8)

Here's how you figure out the formula for a line that goes through two points (x1,y1) and (x2,y2).

The formula with a line that has a slope of m and a y-intercept (the value of y where the line crosses the y-axis) of b is

y = mx + b

The slope, m, of the line is the change in y divided by the change in x.

m = (y2-y1) / (x2-x1)

Now, to find the y-intercept, b, use one of the points to get the x and y values, and solve for b.

y1 = mx1 + b
b = y1 - mx1

Now, let's apply this information to the problem at hand.

x1 = 2,  y1 = 6.
x2 = 6,  y2 = 8

m = (8-6) / (6 - 2)    -- the difference of the y's over the difference of the x's
m = 2/4 = 1/2

b = 6 - (1/2)(2)
b = 6 - 1
b = 5

Now that we know the slope, m = 1/2, and the y-intercept, b = 5, we can write the equation of the line:

y = (1/2)x + 5, which can be written
y = x/2 + 5

Now are we done?  Not yet -- we always check our answers.  Make sure both points (2,6) and (6,8) are on the equation.

Does 6 = 2/2 + 5?  Yes.  6 = 1 + 5.

Does 8 = 6/2 + 5?  Yes.  8 = 3 + 5.

We're done!

2. Without graphing tell whether the graphs of each pair of equations are parallel:
y+5=3x
5x-y=-11

To answer this one, put both equations in slope-intercept form, which is

y = mx + b

In other words, solve each one for y.

y + 5 = 3x
y = 3x - 5

5x - y = -11
-y = 5x - 11
y = -5x + 11

Now look at the slope, which is the x-coefficient, or m, of each of these two equations.  If they're equal, then either the lines are parallel, or they are the same line (which is the case if the y-intercept values are the same).  If the slopes are not equal, then the lines are not parallel.  In this case, the slope of the first line is 3, and the slope of the second line is -5.  They're not equal.  The lines are not parallel.

3. Write an equation of the line containing the specified point and parallel to the indicated line:
(4,0), 4x-y=7

We need to find the slope of the line, so we can find a line with the same slope (that means parallel) that passes through the point.  To find the slope, put the equation in slope-intercept form, which means: solve for y.

4x - y = 7
-y = -4x + 7
y = 4x - 7

Now the x-coefficient is the slope, which is 4.  We can say m=4, where m is the slope.  Now we need to find a the y-intercept of the line with this slope that passes through the point (4,0).  To find the y-intercept, b, use the point's x and y values, and solve for b.

y1 = mx1 + b
b = y1 - mx1

For our problem, x1 = 4, y1 = 0, and m = 4.  Plugging those values in our formula for b, we get

b = 0 - (4)(4), which is -16

Now that we know the slope, m=4, and the y-intercept, b=-16, the equation of the line is

y = 4x - 16

Now are we done?  Not yet -- we always check our answers.  Make sure the points (4,0) is on the equation.

Does 0 = (4)(4) - 16?  Yes.  0 = 16 - 16.

We're done!

4. Write an equation of the line containing the specified point and perpendicular to the indicated line:
(1,8), 5x+2y=7

If the slope of a line is m, then the perpendicular line's slope is -1/m.  To understand this, graph a few lines, and then graph their perpendiculars.  Count the change in y and the change in x of the original line, and then notice that the perpendicular line has the same change in x that the original one had in y, and vice versa, but that either the x or the y (but not both) will change in the opposite direction.  If you didn't get that, don't worry.  You can just trust the rule, if you like.

So let's put this equation in slope-intercept form, and after that, we'll find the slope of a perpendicular line.

5x + 2y = 7
2y = -5x + 7
y = (-5/2)x + 7/2

Now we see the slope of the original line is m = -5/2.  The perpendicular line has a slope  mp-1/m = 2/5.  At this point, we have another problem in which we have a point and a slope, and we need to come up with an equation for a line with that slope that passes through the point.

We can say m=2/5, where m is the slope.  Now we need to find a the y-intercept of the line with this slope that passes through the point (1,8).  To find the y-intercept, b, use the point's x and y values, and solve for b.

y1 = mx1 + b
b = y1 - mx1

For our problem, x1 = 1, y1 = 8, and m = 2/5.  Plugging those values in our formula for b, we get

b = 8 - (2/5)(1), which is 7 3/5, or 38/5.

Now that we know the slope, m=2/5, and the y-intercept, b= 38/5, the equation of the line is

y = (2/5)x + 38/5

Now are we done?  Not yet -- we always check our answers.  Make sure the points (1,8) is on the equation.

Does 8 = (2/5)(1) +  38/5?  Yes.  8  =  2/5 + 38/5  =  40/5

We're done!

### Related Pages in this website

Back to Geometry and Trig

Points and Triangles -- answers the question: is the point inside the triangle?

Using the determinant of a matrix to calculate the area of a polygon

Triangle Area using Determinant

Triangle Area using Vectors

The webmaster and author of this Math Help site is Graeme McRae.