The length of the horizontal leg of the triangle is the distance from the origin  to , which is 4.

The area of a right triangle is half the product of the lengths of its legs  and , so, setting  and  and solving for :

Since this is the vertical distance from the origin, this is also the absolute value of the -coordinate of the -intercept of the line; also, this point is along the positive -axis. The line has -intercept .

The slope of a line, given the intercepts , is 

,

Substitute  and :

. 

To find k, first identify what is known.

The origin means the point (0,0).

Therefore the line passes through the points:

Since a line by definition has the same slope between the points, calculation the slope between the origin and each of the points.

Let

therefore the slope is,

Now let,

thus the slope is,

From here set the slopes equal to each other and solve for k.

Multiply by k on both sides.

The length of the vertical leg of the triangle is the distance from the origin  to , which is 12.

The area of a right triangle is half the product of the lengths of its legs  and , so, setting  and  and solving for :

Since this is the positive horizontal distance from the origin, this is also the -coordinate of the -intercept of the line - that is, the line has -intercept .

The slope of a line, given the intercepts , is 

,

Substitute  and :

While solving the problem requires the same method as the ones above, this is one is more complicated because of the more complex given equations. Start of by deriving a substitute for one of the unknowns. From the second equation we can derive y=75-(5x/2). Since 2y = 150 -5x, we divide both sides by two and find our substitution for y. Then we enter this into the first equation. We now have –x-4(75-(5x/2))=245. Distribute the 4. So we get –x – 300 + 10x = 245. So 9x =545, and x=545/9. Use this value for x and solve for y. The graph below illustrates the solution.

This one is a basic problem with two unknowns in two equations. Derive y=x+10 from the second equation and replace the y in first equation to solve the problem. So, x+10+5x=40 and x = 5. X-y= -10 so y=15. The graph below illustrates the solution.

Again the same process is required. This problem however involved multiplying x by y so is a bit different. We end up with two possible solutions. Derive y=x+1 and solve in the same manner as the ones above. The graph below illustrates the solution.

Similar problem to the one before, with x being divided by y instead of multiplied. Solve in the same manner but keep in mind the way that x/y is graphed. We end up solving for one solution. The graph below illustrates the solution,

Straightforward problem that presents two unknowns with two equations. The student will need to deal with the fractions correctly to get this one right. Other than the fraction the problem is solved in the exact same manner as the rest in this set. The graph below illustrates the solution.

First we need to find the slope. Slope (m) = (y₂ - y₁)/(x₂ - x₁). Substituting in our values (-5 - 4)/(2 - (-1)) = -9/3 = -3 so slope = -3. The formula for a line is y = mx +b. We know m = -3 so now we can pick one of the two points, substitute in the values for x and y, and find b. 4 = (-3)(-1) + b so b = 1. Our formula is thus y = -3x + 1

We will find the equation using slope intercept form: y=mx+b

1. Use the two points to find the slope.

The equation to find the slope of this line using two points is:

  Therefore, m = ¹/₂, so the slope of this line is ¹/₂.

2. Now that we have the slope, we can use one of the points that were given to find the y intercept. In order to do this, substitute y for the y value of the point, and substitute x for the x value of the point.

Using the point (–2, –2), we now have: –2 = (¹/₂)(–2) + b.

Simplify the equation to solve for b. b = –1

3. In this line m = ¹/₂ and b = –1

4. Therefore, y = ¹/₂x –1