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Tutorial 1.15 Write the standard form of an equation of a line given its slope and a point
Although the slope-intercept (y = mx + b) form is used most often when using graphing technology, there may be times when we want to express the desired equation in its standard (Ax + By = C) form.
We can accomplish this desired format in one of three ways:
• Reformat the slope-intercept (y = mx + b) form into standard (Ax + By = C) form.
• Reformat the point-slope (y2 – y1 = m(x2 – x1)) form into standard (Ax + By = C) form.
• Use the slope and some arithmetic to determine the standard (Ax + By = C) form.
example 1.15a Write a slope-intercept equation for the line that has the given slope and passes through the given point: slope = –2, P = (0, 3)
method #1: reformatting the slope-intercept form…
Since all of the values in the slope-intercept (y = mx + b) form of the equation are integers, just rearrange the terms so that the x and y terms are on the left-hand side of the equation while the non-variable term is on the right-hand side.
slope = –2 and P = (0, 3) → slope-intercept: y = –2x + 3 Adding 2x to both sides of the equation, we get: 2x + y = 3 method #2: reformatting the point-slope form…
1. Use the point-slope format to structure an equation:
y – y1 = m(x – x1) → y – 3 = –2(x – 0) → y = –2(x – 0) + 3 = –2x + 0 + 3 = –2x + 3 → y = –2x + 3 Adding 2x to both sides of the equation, we get: 2x + y = 3
method #3: using the slope…
Since the slope of any linear equation is –A/B and the slope for this example is –2, we know that A = 2 when B = 1. Since the given point is (0, 3), we know that x = 0 when y = 3. Using all of this information, we can determine the value of C in the standard (Ax + By = C) form of the desired equation:
(A)(x) + (B)(y) = C → C = (2)(0) + (1)(3) = 0 + 3 = 3 → equation is: 2x + y = 3
example 1.15b Write a slope-intercept equation for the line that has the given slope and passes through the given point: slope = 34, P = (0, –5)
method #1: reformatting the slope-intercept form…
The slope and the y-intercept is given:
slope = 34and P = (0, –5) → slope-intercept: y = (34)(x) – 5
Since the slope value in the slope-intercept (y = mx + b) form of the equation is not an integer, we need to multiply the entire equation through by the denominator of the slope…for this example, that would be 4. After multiplying both sides of the equation by 4, we get the transformed equation: 4y = 3x – 20.
We can now rearrange the terms so that the x and y terms are on the left-hand side of the equation while the non-variable term is on the right-hand side. Subtracting 3x from both sides of the transformed equation, we get: –3x + 4y = –20. However, recall that in standard form, the value of A should be positive…therefore, we need to multiply the entire equation by –1…which results in: 3x – 4y = 20.
method #2: eformatting the point-slope form…
1. Use the point-slope format to structure an equation:
y – y1 = m(x – x1) → y – (–5) = (34)(x – 0) → y + 5 = (34)(x – 0) y = (34)(x – 0) – 5 → y = (34)(x) – 5
2. Multiply both sides of the equation by the denominator (4): 4y = 3x – 20 3. Subtract 4y from both sides of the equation: 0 = 3x – 4y – 20
4. Add 20 to both sides of the equation: 3x – 4y = 20 method #3: using the slope…
Since the slope of any linear equation is –A/B and the slope for this example is 34, we know that A = 3 when B = –4. (NOTE: Since the A value needs to be positive, make it positive now rather than later.) Since the given point is (0, –5), we know that x = 0 when y = –5. Using all of this information, we can determine the value of C in the standard (Ax + By = C) form of the desired equation:
(A)(x) + (B)(y) = C → C = (3)(0) + (–4)(–5) = 0 + 20 = 20 → equation: 3x + 4y = 20
example 1.15c Write a slope-intercept equation for the line that has the given slope and passes through the given point: slope = 23, P = (3, 2)
method #1: reformatting the slope-intercept form…
slope =23and P = (3, 2) → y = (23)(x) + b → 2 = (23)(3) + b
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Since the slope value in the slope-intercept (y = mx + b) form of the equation is not an integer, we need to multiply the entire equation through by the denominator of the slope…for this example, that would be 3. After multiplying both sides of the equation by 3, we get the transformed equation: 3y = 2x. We can now rearrange the terms so that the x and y terms are on the left-hand side of the equation while the non-variable term is on the right-hand side. Since we want 2x to be positive, subtract 3x from both sides of the transformed equation, to get: 0 = 2x – 3y (which can also be written as: 2x – 3y = 0).
method #2: reformatting the point-slope form…
1. Use the point-slope format to structure an equation:
y – y1 = m(x – x1) → y – 2 = (23)(x – 3) → y = (23)(x – 3) + 2 y = (233)(x) – 2 + 2 → y = (23)(x) + 0 → y = (23)(x)
2. Multiply both sides of the equation by the denominator (3): 3y = 2x 3. Subtract 3y from both sides of the equation: 0 = 2x – 3y
method #3: using the slope…
Since the slope of any linear equation is –A/B and the slope for this example is 2/3, we know that A = 2 when B = –3. Since the given point is (3, 2), we know that x = 3 when y = 2. Using all of this information, we can determine the value of C in the standard (Ax + By = C) form of the desired equation:
(A)(x) + (B)(y) = C → C = (2)(3) + (–3)(2) = 6 + –6 = 0 → equation: 2x – 3y = 0
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