logo for solving-math-problems.com
leftimage for solving-math-problems.com

College Algebra - Standard Form of the Equation










































1) Find the standard form of the equation of the circle with endpoints of a diameter at the points (1,6) and (-5,4).

2) Find the standard form of the equation of the circle with endpoints of a diameter at the points (5,4) and (-9,7).

3) Use the given conditions to write an equation for the line in slope intercept form. Passing through (3,-6) and perpendicular to the line whose equation is 2x-3y=5. Write an equation for the line in slope intercept form.

4) Find the equation of the line passing through the points (1,1) and (-2,4). Write the equation in point slope form, slope intercept form, and standard form.

Comments for College Algebra - Standard Form of the Equation

Click here to add your own comments

Oct 07, 2012
Standard Form of the Equation
by: Staff


Answer:

Part I


1) Find the standard form of the equation of the circle with endpoints of a diameter at the points (1,6) and (-5,4).

The equation for a circle with radius “r” and center coordinates of (h,k) is:

(x - h)² + (y - k)² = r²

The line segment (-5,4) to (1,6) represents the diameter of the circle.

The coordinates for the center of the circle is the point half way between (1,6) and (-5,4).



The midpoint between two points (called the Midpoint Formula) is given by:



x1 + x2   y1 + y2

------- , -------
2 2



Calculate the midpoint between points (-5,4) and (1,6)



-5 + 1     4 + 6

------- , -------
2 2



-4 10
------- , -------
2 2



(-2 , 5)



The radius of the circle is the distance between the center of the circle and one of the points on the circumference of the circle:


     _______________________

r = √(x2 - x1)² + (y2 - y1





Calculate the radius using points (-2 , 5) and (1,6)


     _______________________

r = √(1 – (-2))² + (6 - 5)²

___________________
r = √(1 + 2)² + (6 - 5)²

___________
r = √(3)² + (1)²

_____
r = √9 + 1

__
r = √10


r ≈ 3.1622776601684



----------------------------------------------------

Oct 07, 2012
Standard Form of the Equation
by: Staff


----------------------------------------------------
Part II


Since the Standard Form for the equation of a circle with radius “r” and center coordinates of (h,k) is:


(x - h)² + (y - k)² = r²




The The Standard Form for the equation of a circle with a radius of “r = √10” and center coordinates of (-2 , 5) is:


(x - (-2))² + (y - 5)² = 10


(x + 2)² + (y - 5)² = 10


Final Answer:


                 (x + 2)² + (y - 5)² = 10


 Circle Graph:  (x + 2)² + (y - 5)² = 10










---------------------------------------------------


2) Find the standard form of the equation of the circle with endpoints of a diameter at the points (5,4) and (-9,7).


The equation for a circle with radius “r” and center coordinates of (h,k) is:

(x - h)² + (y - k)² = r²

The line segment (-9,7) to (5,4) represents the diameter of the circle.

The coordinates for the center of the circle is the point half way between (-9,7) and (5,4).

----------------------------------------------------

Oct 07, 2012
Standard Form of the Equation
by: Staff

----------------------------------------------------
Part III


The midpoint between two points (called the Midpoint Formula) is given by:



x1 + x2   y1 + y2
------- , -------
2 2



Calculate the midpoint between points (-9,7) and (5,4)



-9 + 5     7 + 4
------- , -------
2 2



-4 11
------- , -------
2 2



(-2 , 5.5)



The radius of the circle is the distance between the center of the circle and one of the points on the circumference of the circle:


     _______________________
r = √(x2 - x1)² + (y2 - y1





Calculate the radius using points (-2,5.5) and (5,4)


     ________________________
r = √(5 – (-2))² + (4 - 5.5)²

_____________________
r = √(5 + 2)² + (4 - 5.5)²

______________
r = √(7)² + (-1.5)²

_________
r = √49 + 2.25

_____
r = √51.25


r ≈ 7.1589105316382


Since the Standard Form for the equation of a circle with radius “r” and center coordinates of (h,k) is:


(x - h)² + (y - k)² = r²

----------------------------------------------------

Oct 07, 2012
Standard Form of the Equation
by: Staff

----------------------------------------------------
Part IV


The The Standard Form for the equation of a circle with a radius of “r = √51.25” and center coordinates of (-2 , 5.5) is:


(x - (-2))² + (y - 5.5)² = 51.25


(x + 2)² + (y - 5.5)² = 51.25


Final Answer:


                 (x + 2)² + (y - 5.5)² = 51.25


 Circle Graph:  (x + 2)² + (y - 5.5)² = 51.25






---------------------------------------------------


3) Use the given conditions to write an equation for the line in slope intercept form. Passing through (3,-6) and perpendicular to the line whose equation is 2x-3y=5. Write an equation for the line in slope intercept form.

The Step 1: Solve for the Slope Intercept Form of 2x - 3y = 5

y = mx + b


2x - 3y = 5

2x - 3y + 3y = 5 + 3y

2x + 0 = 5 + 3y

2x = 5 + 3y

2x - 5 = 5 + 3y - 5

2x - 5 = 5 - 5 + 3y

2x - 5 = 0 + 3y

2x - 5 = 3y

3y = 2x - 5

3y / 3 = (2x - 5) / 3

y * (3 / 3) = (2x - 5) / 3

y * (1) = (2x - 5) / 3

y = (2x - 5) / 3

y = (2/3)x - 5/3

----------------------------------------------------

Oct 07, 2012
Standard Form of the Equation
by: Staff

----------------------------------------------------
Part V


Step 2: Determine the slope of the equation and the slope of the perpendicular equation


Slope of the equation

m1 = 2/3

slope of equation perpendicular to this curve:

m2 = -3/2


Step 3: Determine the equation for the perpendicular curve


equation for perpendicular line:

y = (-3/2)x + b

The new curve must pass through point (3,-6).

Use this information to solve for “b”

y = (-3/2)x + b

-6 = (-3/2)(3) + b

-6 = (-9/2) + b

-6 + (9/2) = (-9/2) + b + (9/2)

-6 + (9/2) = (-9/2) + (9/2) + b

-6 + (9/2) = 0 + b

-6 + (9/2) = b

b = -6 + (9/2)

b = -6 + 4.5

b = -1.5


The slope intercept form of the new equation is:

y = (-3/2)x - 1.5

Perpendicular Line Graph




----------------------------------------------------

Oct 07, 2012
Standard Form of the Equation
by: Staff

----------------------------------------------------
Part VI


4) Find the equation of the line passing through the points (1,1) and (-2,4). Write the equation in point slope form, slope intercept form, and standard form.


Step 1: Solve for the Slope of the line passing through the two points (1,1) and (-2,4)

slope = “rise” over “run”, 
or “rise” DIVIDED BY “run”

slope = (change in “y” values)/(change in “x” values)

slope = (y₂ - y₁)/(x₂ - x₁)


The x-y coordinates of the two points are:
(1,1) and (-2,4).


Note: These points are not listed in the
proper order (left to right)

They should be listed (-2,4) and (1,1) because
the slope will be calculated as the change of
“y” when x increases. In this case “x” will
increase from -2 to 1 (left to right).

(x₁,y₁)= (-2,4)

(x₂,y₂)=(1,1)

Points

Left and Right

(-2,4) and (1,1)

(x₁, y₁) and (x₂,y₂)

x₁ = -2
x₂ = 1

y₁ = 4
y₂ = 1

slope = (y₂ - y₁)/(x₂ - x₁)

slope = (1 - 4)/( 1 - (-2))

slope = (-3)/( 1 + 2)

slope = (-3)/( 3)

Slope = -1

m (slope) = -1


The Point-Slope Form

Point-Slope Form:

(y - y₁) = m (x - x₁)

this format uses a single known point
(x₁,y₁) and the
slope m (which is also a known value)

(y - 4) = -1 (x - (-2))

(y - 4) = -1 (x + 2)



The Slope Intercept Form

Slope Intercept Form

y = mx + b

(y - 4) = -1 (x + 2)

y - 4 = -1x - 2

y - 4 + 4 = -1x - 2 + 4

y + 0 = -1x - 2 + 4

y = -1x - 2 + 4

y = -1x + 2


----------------------------------------------------

Oct 07, 2012
Standard Form of the Equation
by: Staff

----------------------------------------------------
Part VII

The Standard Form

Standard Form

Ax + By = C

A, B, and C are constants

y = -1x + 2

y + 1x = -1x + 2 + 1x

y + 1x = -1x + 1x + 2

y + 1x = 0 + 2

y + 1x = 2

x + y = 2


Equation of the line passing through the points (1,1) and (-2,4)









Thanks for writing.

Staff
www.solving-math-problems.com


Click here to add your own comments

Join in and write your own page! It's easy to do. How? Simply click here to return to Math Questions & Comments - 01.



Copyright © 2008-2014. All rights reserved. Solving-Math-Problems.com