Question

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)

Answer 1

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. 

Answer 2

Part 1)
`7x-4y=42`

Part 2)
`9x+5y=-45`

Part 3)
`6x+7y=14`

part 4)
`5x+4y=20`

Explanation:

we know that

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

`Ax + By = C`

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
`(6,0)` and
`(2,-7)`

we know that

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

`m=(y2-y1)/(x2-x1)`

Substitute the values

`m=(-7-0)/(2-6)`

`m=(-7)/(-4)`

`m=(7)/(4)`

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

`y-y1=m(x-x1)`

we have

`m=(7)/(4)`

`Point (6,0)`

substitute

`y-0=(7)/(4)(x-6)`

`y=(7)/(4)x-(42)/(4)`

Multiply by
`4` both sides

`4y=7x-42`

convert in standard form

`7x-4y=42`

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

we know that

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

`m=(y2-y1)/(x2-x1)`

Substitute the values

`m=(-9-0)/(0+5)`

`m=-(9)/(5)`

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

`y-y1=m(x-x1)`

we have

`m=-(9)/(5)`

`Point (-5,0)`

substitute

`y-0=-(9)/(5)(x+5)`

`y=-(9)/(5)x-9`

Multiply by
`5` both sides

`5y=-9x-45`

convert in standard form

`9x+5y=-45`

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

we know that

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

`m=(y2-y1)/(x2-x1)`

Substitute the values

`m=(2-8)/(0+7)`

`m=-(6)/(7)`

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

`y-y1=m(x-x1)`

we have

`m=-(6)/(7)`

`Point (0,2)`

substitute

`y-2=-(6)/(7)(x-0)`

`y=-(6)/(7)x+2`

Multiply by
`7` both sides

`7y=-6x+14`

convert in standard form

`6x+7y=14`

Part 4) Determine the standard form of the equation of the line that passes through
`(0,5)` and
`(4,0)`

we know that

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

`m=(y2-y1)/(x2-x1)`

Substitute the values

`m=(0-5)/(4-0)`

`m=-(5)/(4)`

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

`y-y1=m(x-x1)`

we have

`m=-(5)/(4)`

`Point (0,5)`

substitute

`y-5=-(5)/(4)(x-0)`

`y=-(5)/(4)x+5`

Multiply by
`4` both sides

`4y=-5x+20`

convert in standard form

`5x+4y=20`