Question

Write a linear equation in standard form for the line that goes through

(-14,-5) and (4,7)
A. Y-5. }(x-4)
O B. -2x+ 3y = 1
C. - 2x + 3y - 13
O D. 2x - 3y = 13

Answer 1

`3y - 2x = 13`

Explanation:

Given

Points (-14,-5) and (4,7)

Required

Find its linear equation in a standard form

To find the linear form, we start by calculating the slope of the line

This is calculated as thus:

`m = (y_2 - y_1)/(x_2 - x_1)`

Where
`(x_1,y_1) = (-14,-5)\ and\ (x_2,y_2) = (4,7)`

So,
`m = (y_2 - y_1)/(x_2 - x_1)` becomes

`m = (7 - (-5))/(4 - (-14))`

`m = (7 + 5)/(4 + 14)`

`m = (12)/(18)`

Simplify fraction to lowest term

`m = (6*2)/(6*3)`

`m = (2)/(3)`

The equation of the line can then be calculated using any of the given points;

Using

`m = (y - y_1)/(x - x_1)`

`Where\ (x_1,y_1) = (-14,-5)\ and\ m = (2)/(3)`

We have

`(2)/(3) = (y-(-5))/(x-(-14))`

`(2)/(3) = (y+5)/(x+14)`

Multiply both sides by 3

`3 * (2)/(3) = (y+5)/(x+14) * 3`

`2 = (y+5)/(x+14) * 3`

`2 = (3(y+5))/(x+14)`

Multiply both sides by x + 14

`2 * (x + 14) = (3(y+5))/(x+14) * (x + 14)`

`2 * (x + 14) = 3(y+5)`

Open brackets

`2 * x + 2 * 14 = 3* y+ 3 * 5`

`2x + 28 = 3y+ 15`

Subtract 2x from both sides

`2x - 2x + 28 = 3y+ 15 - 2x`

`28 = 3y+ 15 - 2x`

Subtract 15 from both sides

`28 - 15 = 3y+ 15 - 2x - 15`

`28 - 15 = 3y - 2x + 15- 15`

`13 = 3y - 2x`

Reorder

`3y - 2x = 13`

Hence, the equation of the line in standard form is
`3y - 2x = 13`