How to find the slope of perpendicular lines - PSAT Math

The slope of line m is the (y2 - y1) / (x2 - x1) = (5-7) / (-2 - 3)

= -2 / -5

= 2/5

To find the slope of a line perpendicular to a given line, we must take the negative reciprocal of the slope of the given line.

Thus the slope of line k is the negative reciprocal of 2/5 (slope of line m), which is -5/2.

First, solve for the equation of the line in the form of y = mx + b so that you can determine the slope, m of the line:

8x + 16y = 48

16y = -8x + 48

y = -(8/16)x + 48/16

y = -(1/2)x + 3

Therefore the slope (or m) = -1/2

The slope of a perpendicular line is the negative inverse of the slope.

m = - (-2/1) = 2

This is a system of equations problem with an x squared, to be solved just like the rest of the problem set. Two solutions are required due to the x2. The graph below illustrates those solutions.

First we find the slope of line M by using the slope formula (_y_2 – _y_1)/(_x_2 – _x_1).

(–5 – 2)/(3 – 2) = –7/1. This means the slope of Line M is –7. A line perpendicular to Line M will have a negative reciprocal slope. Thus, the answer is y = (1/7)x + 3.

Solving the equation follows the same system as the first problem. However since x is squared in this problem we will have two possible solutions for each unknown. Again substitute y=11-x and solve from there. Hence, x2+11-x=31. So x2-x=20. 5 squared is 25, minus 5 is 20. Now we know 5 is one of our solutions. Then we must solve for the second solution which is -4. -4 squared is 16 and 16 –(-4) is 20. The last step is to solve for y for the two possible solutions of x. We get 15 and 6. The graph below illustrates to solutions.

This problem is very similar to number 2. Derive y=14-x and solve from there. The graph below illustrates the solution.

The problem involves the same method used for the rest of the practice set. However since the x is squared we will have multiple solutions. Solve this one in the same way as number 2. However be careful to notice that the y value is the same for both x values. The graph below illustrates the solution.

We know that the line passes through point B, but we must calculate its slope in order to find the equation that defines the line. Because the line is tangent to the circle, it must make a right angle with the radius of the circle at point B. Therefore, the slope of the line is perpendicular to the slope of the radius that connects the center of the circle to point B. First, we can find the slope of the radius, and then we can determine the perpendicular slope.

The radius passes through points C and B. We can use the formula for the slope (represented as ) between two points to find the slope of the radius.

Point C: (2,-5) and point B: (7,-3)

This is the slope of the radius, but we need to find the slope of the line that is perpendicular to the radius. This value will be equal to the negative reciprocal.

Now we know the slope of the tangent line. We can use the point-slope formula to find the equation of the line. The formula is shown below.

Plug in the give point that lies on the tangent line (point B) and simplify the equation.

Multiply both sides by two in order to remove the fraction.

Distribute both sides.

Add to both sides.

Subtract six from both sides.

The answer is .

Looking at the graph, we can tell the slope of the line is 3 with a -intercept of , so the equation of the line is:

y=3x-4

A perpendicular line to this would have a slope of -$\frac{1}{3}$, and would pass through the point so it follows:

y=-$\frac{1}{3}$x+crightarrow 2=-$\frac{1}{3}$(2)+crightarrow c=8/3rightarrow y=-$\frac{1}{3}$x+\frac{8}{3}$

First we will calculate the slope as follows:

m=\frac{y_2-y_1}{x_2-x_1}$=\frac{7-5}{4-3}$=\frac{2}{1}$=2

And our equation for a line is

y=mx+b=2x+b

Now we need to calculate b. We can pick either of the points given and solve for b

5=2(3)+b

b=-1

Our equation for the line becomes

y=2x-1

In order to move from the upper left point to the lower right point, it is necessary to move down 3 units and right six units. This is a rise of and a run of 6. The slope of a line is the ratio of rise to run, so slope of the line shown is .

A line perpendicular to this will have a slope equal to the opposite of the reciprocal of . This is .

The slope of line m is the (y2 - y1) / (x2 - x1) = (5-7) / (-2 - 3)

= -2 / -5

= 2/5

To find the slope of a line perpendicular to a given line, we must take the negative reciprocal of the slope of the given line.

Thus the slope of line k is the negative reciprocal of 2/5 (slope of line m), which is -5/2.

First, solve for the equation of the line in the form of y = mx + b so that you can determine the slope, m of the line:

8x + 16y = 48

16y = -8x + 48

y = -(8/16)x + 48/16

y = -(1/2)x + 3

Therefore the slope (or m) = -1/2

The slope of a perpendicular line is the negative inverse of the slope.

m = - (-2/1) = 2

This is a system of equations problem with an x squared, to be solved just like the rest of the problem set. Two solutions are required due to the x2. The graph below illustrates those solutions.

First we find the slope of line M by using the slope formula (_y_2 – _y_1)/(_x_2 – _x_1).

(–5 – 2)/(3 – 2) = –7/1. This means the slope of Line M is –7. A line perpendicular to Line M will have a negative reciprocal slope. Thus, the answer is y = (1/7)x + 3.

Solving the equation follows the same system as the first problem. However since x is squared in this problem we will have two possible solutions for each unknown. Again substitute y=11-x and solve from there. Hence, x2+11-x=31. So x2-x=20. 5 squared is 25, minus 5 is 20. Now we know 5 is one of our solutions. Then we must solve for the second solution which is -4. -4 squared is 16 and 16 –(-4) is 20. The last step is to solve for y for the two possible solutions of x. We get 15 and 6. The graph below illustrates to solutions.

This problem is very similar to number 2. Derive y=14-x and solve from there. The graph below illustrates the solution.

The problem involves the same method used for the rest of the practice set. However since the x is squared we will have multiple solutions. Solve this one in the same way as number 2. However be careful to notice that the y value is the same for both x values. The graph below illustrates the solution.

We know that the line passes through point B, but we must calculate its slope in order to find the equation that defines the line. Because the line is tangent to the circle, it must make a right angle with the radius of the circle at point B. Therefore, the slope of the line is perpendicular to the slope of the radius that connects the center of the circle to point B. First, we can find the slope of the radius, and then we can determine the perpendicular slope.

The radius passes through points C and B. We can use the formula for the slope (represented as ) between two points to find the slope of the radius.

Point C: (2,-5) and point B: (7,-3)

This is the slope of the radius, but we need to find the slope of the line that is perpendicular to the radius. This value will be equal to the negative reciprocal.

Now we know the slope of the tangent line. We can use the point-slope formula to find the equation of the line. The formula is shown below.

Plug in the give point that lies on the tangent line (point B) and simplify the equation.

Multiply both sides by two in order to remove the fraction.

Distribute both sides.

Add to both sides.

Subtract six from both sides.

The answer is .

Looking at the graph, we can tell the slope of the line is 3 with a -intercept of , so the equation of the line is:

y=3x-4

A perpendicular line to this would have a slope of -$\frac{1}{3}$, and would pass through the point so it follows:

y=-$\frac{1}{3}$x+crightarrow 2=-$\frac{1}{3}$(2)+crightarrow c=8/3rightarrow y=-$\frac{1}{3}$x+\frac{8}{3}$