Let and

The slope is geven by: so

Then we use the slope-intercept form of an equation; so

And we convert

to standard form.

In order to find the equation of the line, we need to find two points on the line. We are told that the line passes through the y-intercept and one x-intercept of y = –x_2 – 2_x + 8.

First, let's find the y-intercept, which occurs where x = 0. We can substitute x = 0 into our equation for y.

y = –(0)2 – 2(0) + 8 = 8

The y-intercept occurs at (0,8).

To determine the x-intercepts, we can set y = 0 and solve for x.

0 = –x_2 – 2_x + 8

–x_2 – 2_x + 8 = 0

Multiply both sides by –1 to minimize the number of negative coefficients.

x_2 + 2_x – 8 = 0

We can factor this by thinking of two numbers that multiply to give us –8 and add to give us 2. Those numbers are 4 and –2.

x_2 + 2_x – 8= (x + 4)(x – 2) = 0

Set each factor equal to zero.

x + 4 = 0

Subtract 4.

x = –4

Now set x – 2 = 0. Add 2 to both sides.

x = 2

The x-intercepts are (–4,0) and (2,0).

However, we don't know which x-intercept the line passes through. But, we are told that the line has a negative slope. This means it must pass through (2,0).

The line passes through (0,8) and (2,0).

We can use slope-intercept form to write the equation of the line. According to slope-intercept form, y = mx + b, where m is the slope, and b is the y-intercept. We already know that b = 8, since the y-intercept is at (0,8). Now, all we need is the slope, which we can find by using the following formula:

m = (0 – 8)/(2 – 0) = –8/2 = –4

y = mx + b = –4_x_ + 8

The answer is y = –4_x_ + 8.

First, we need to find the slope of the above line.

The slope of a line. given two points can be calculated using the slope formula

Set :

The slope of a line perpendicular to it has as its slope the opposite of the reciprocal of 3, which is .

Since we want the line to have the same -intercept as the above line, which is the point , we can use the slope-intercept form to help us. We set

, and solve for in:

Substitute for and in the slope-intercept form, and the equation is

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, solve for slope.

Then, substitute one of the points into the equation y=mx+b.

This leaves us with the equation

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

Let P1 (1, 1) and P2 (–2, 3).

First, find the slope using m = rise ÷ run = (y2 – y1)/(x2 – x1) giving m = –2/3.

Second, substutite the slope and a point into the slope-intercept equation y = mx + b and solve for b giving b = 5/3.

Third, convert the slope-intercept form into the standard form giving 2x + 3y = 5.

In order to find the equation of the line, we will first need to find the slope between the two points through which it passes. The slope, , of a line that passes through the points and is given by the formula below:

We are given our two points, (4,7) and (8,10), allowing us to calculate the slope.

Next, we can use point slope form to find the equation of a line with this slope that passes through one of the given points. We will use (4,7).

Multiply both sides by four to eliminate the fraction, and simplify by distribution.

Subtract from both sides and add twelve to both sides.

This gives our final answer:

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.

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.