  # 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

 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)² ``` 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  --------------------------------------------------- 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 IIIThe 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)²` 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  --------------------------------------------------- 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 VStep 2: Determine the slope of the equation and the slope of the perpendicular equation`Slope of the equationm1 = 2/3slope 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 + bThe 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 = -1.5The slope intercept form of the new equation is:y = (-3/2)x - 1.5`  ----------------------------------------------------

 Oct 07, 2012 Standard Form of the Equation by: Staff ----------------------------------------------------Part VI4) 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 Formy = mx + b(y - 4) = -1 (x + 2) y - 4 = -1x - 2 y - 4 + 4 = -1x - 2 + 4y + 0 = -1x - 2 + 4y = -1x - 2 + 4y = -1x + 2` ----------------------------------------------------

 Oct 07, 2012 Standard Form of the Equation by: Staff ----------------------------------------------------Part VIIThe Standard Form `Standard FormAx + By = C A, B, and C are constantsy = -1x + 2y + 1x = -1x + 2 + 1xy + 1x = -1x + 1x + 2 y + 1x = 0 + 2 y + 1x = 2 x + y = 2 `  Thanks for writing. Staff www.solving-math-problems.com