Question

Can someone help me with these ones :(? (with full process)

1.) (3,2) and (4,j),m=1

2). (5,0) and (1,k),m=1/2

3).(x, 2) and (3, -4), m = 2

4). (12, -4) y (r, 2), m = -1/2

Answer 1

The answers for:

1. j = 3
2. k = -2
3. x = 6
4. r = 0

Explanation:

In order to find the value of expression, you have to apply gradient formula :

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

So for Question 1,

`let \: (3,2) \: be \: (x1,y1) \\ let \: (4,j) \: be \: (x2,y2) \\ let \: m = 1`

`(j - 2)/(4 - 3) = 1`

`(j - 2)/(1) = 1`

`j - 2 = 1`

`j = 1 + 2 = 3`

Question 2,

`let \: (5,0) \: be \: (x1,y1) \\ let \: (1,k) \: be \: (x2,y2) \\ let \: m = (1)/(2)`

`(k - 0)/(1 - 5) = (1)/(2)`

`(k)/( - 4) = (1)/(2)`

`k = (1)/(2) * - 4 = - 2`

Question 3,

`let \: (x,2) \: be \: (x1,y1) \\ let \: (3, - 4) \: be \: (x2,y2) \\ let \: m = 2`

`( - 4 - 2)/(3 - x) = 2`

`( - 6)/(3 - x ) = 2`

`- 6 = 2(3 - x)`

`- 6 = 6 - 2x`

`- 6 - 6 = - 2x`

`- 2x = - 12`

`x = - 12 / - 2 = 6`

Question 4,

`let \: (12, - 4) \: be \: (x1,y1) \\ let \: (r,2) \: be \: (x2,y2) \\ let \: m = - (1)/(2)`

`(2 - ( - 4))/(r - 12) = - (1)/(2)`

`(6)/(r - 12) = - (1)/(2)`

`- 1(r - 12) = 2(6)`

`- r + 12 = 12`

`r = (12 - 12) / - 1 = 0`