Slope and Line Equations - PSAT Math

The slope of a line is defined as a change in the y coordinates over a change in the x coordinates (rise over run).

To calculate the slope of a line, use the following formula:

The slope of the line that passes these two points are simply ∆y/∆x = (3-5)/(2+3) = -2/5

The slope is negative as it starts in quadrant 2 and ends in quadrant 4. Slope is equivlent to the change in y divided by the change in x. The change in y is greater than the change in x, which implies that the slope must be less than –1, leaving –2 as the only possible solution.

Let and

so the slope becomes .

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

Set :

For each equation, solve for and express in the slope-intercept form . The coefficient of will be the slope.

Slope:

Slope:

Slope:

Slope: .

The line of the equation

is the one with slope 4.

Based on the graph the y-intercept is 4. So we can eliminate choice y = x/2 - 4.

The graph is rising to the right which means our slope is positive, so we can eliminate choice y = -1/2x + 4.

Based on the line, if we start at (0,4) and go up 1 then 2 to the right we will be back on the line, meaning we have a slope of (1/2).

Using the slope intercept formula we can plug in y= (1/2)x + 4.

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

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.

The answer is 22.5.

From the image we can tell that angle a and angle b are complimentary

a + b = 90 and 3a = b

a + 3a = 90

a = 22.5

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.