Determine the standard form of the equation of the line that passes through (6,0) and (2,-7).. . Determine the standard form of the equation of the line that passes through (-5,0) and (0,-9). . Determine the standard form of the equation of the line that passes through (-7,8) and (0,2). . Determine the standard form of the equation of the line that passes through (0,5) and (4,0)

Question

jordayapeycu jordayapeycu

23-06-2016
Mathematics

contestada

Determine the standard form of the equation of the line that passes through (6,0) and (2,-7).. . Determine the standard form of the equation of the line that passes through (-5,0) and (0,-9). . Determine the standard form of the equation of the line that passes through (-7,8) and (0,2). . Determine the standard form of the equation of the line that passes through (0,5) and (4,0)

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meerkat18 meerkat18 · Answer 1 · 2016-07-06T20:40:26+02:00

To answer these problems, we first start finding out the slope. in the first, slope is (-7/-4) or 7/4. we plug in to the equation y-y1 = m(x-x1). hence, y- 0 = 7/4 (x-6). The standard form is 4y = 7x -42. The second one has slope of 9/5. Plugging, we get y-0 = 9/5 (x+5). the standard form then is 5y = 9x +45. The third one's slope is -6/7. Plugging, we get y-2 = -6/7 (x-0) or 7y -14 = -6x or 7y = -6 x +14. Th last one's slope os -5/4. Plugging, we get y-0 = -5/4 (x-4) or equal to 4y = -5x +20.

calculista calculista · Answer 2 · 2019-01-22T21:35:38+01:00

Answer:

Part 1) [tex]7x-4y=42[/tex]

Part 2) [tex]9x+5y=-45[/tex]

Part 3) [tex]6x+7y=14[/tex]

part 4) [tex]5x+4y=20[/tex]

Step-by-step explanation:

we know that

The standard form of the equation of the line is equal to

[tex]Ax + By = C[/tex]

where

A is a positive integer

B, and C are integers

Part 1) Determine the standard form of the equation of the line that passes through [tex](6,0)[/tex] and [tex](2,-7)[/tex]

we know that

The formula to calculate the slope between two points is equal to

[tex]m=\frac{y2-y1}{x2-x1}[/tex]

Substitute the values

[tex]m=\frac{-7-0}{2-6}[/tex]

[tex]m=\frac{-7}{-4}[/tex]

[tex]m=\frac{7}{4}[/tex]

The equation of the line into point slope form is equal to

[tex]y-y1=m(x-x1)[/tex]

we have

[tex]m=\frac{7}{4}[/tex]

[tex]Point (6,0)[/tex]

substitute

[tex]y-0=\frac{7}{4}(x-6)[/tex]

[tex]y=\frac{7}{4}x-\frac{42}{4}[/tex]

Multiply by [tex]4[/tex] both sides

[tex]4y=7x-42[/tex]

convert in standard form

[tex]7x-4y=42[/tex]

Part 2) Determine the standard form of the equation of the line that passes through [tex](-5,0)[/tex] and [tex](0,-9)[/tex]

we know that

The formula to calculate the slope between two points is equal to

[tex]m=\frac{y2-y1}{x2-x1}[/tex]

Substitute the values

[tex]m=\frac{-9-0}{0+5}[/tex]

[tex]m=-\frac{9}{5}[/tex]

The equation of the line into point slope form is equal to

[tex]y-y1=m(x-x1)[/tex]

we have

[tex]m=-\frac{9}{5}[/tex]

[tex]Point (-5,0)[/tex]

substitute

[tex]y-0=-\frac{9}{5}(x+5)[/tex]

[tex]y=-\frac{9}{5}x-9[/tex]

Multiply by [tex]5[/tex] both sides

[tex]5y=-9x-45[/tex]

convert in standard form

[tex]9x+5y=-45[/tex]

Part 3) Determine the standard form of the equation of the line that passes through [tex](-7,8)[/tex] and [tex](0,2)[/tex]

we know that

The formula to calculate the slope between two points is equal to

[tex]m=\frac{y2-y1}{x2-x1}[/tex]

Substitute the values

[tex]m=\frac{2-8}{0+7}[/tex]

[tex]m=-\frac{6}{7}[/tex]

The equation of the line into point slope form is equal to

[tex]y-y1=m(x-x1)[/tex]

we have

[tex]m=-\frac{6}{7}[/tex]

[tex]Point (0,2)[/tex]

substitute

[tex]y-2=-\frac{6}{7}(x-0)[/tex]

[tex]y=-\frac{6}{7}x+2[/tex]

Multiply by [tex]7[/tex] both sides

[tex]7y=-6x+14[/tex]

convert in standard form

[tex]6x+7y=14[/tex]

Part 4) Determine the standard form of the equation of the line that passes through [tex](0,5)[/tex] and [tex](4,0)[/tex]

we know that

The formula to calculate the slope between two points is equal to

[tex]m=\frac{y2-y1}{x2-x1}[/tex]

Substitute the values

[tex]m=\frac{0-5}{4-0}[/tex]

[tex]m=-\frac{5}{4}[/tex]

The equation of the line into point slope form is equal to

[tex]y-y1=m(x-x1)[/tex]

we have

[tex]m=-\frac{5}{4}[/tex]

[tex]Point (0,5)[/tex]

substitute

[tex]y-5=-\frac{5}{4}(x-0)[/tex]

[tex]y=-\frac{5}{4}x+5[/tex]

Multiply by [tex]4[/tex] both sides

[tex]4y=-5x+20[/tex]

convert in standard form

[tex]5x+4y=20[/tex]