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

Answer:

The answers for:

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

Step-by-step explanation:

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

$m = \frac{y2 - y1}{x2 - x1}$

So for Question 1,

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

$\frac{j - 2}{4 - 3} = 1$

$\frac{j - 2}{1} = 1$

$j - 2 = 1$

$j = 1 + 2 = 3$

Question 2,

$let \: (5,0) \: be \: (x1,y1) \\ let \: (1,k) \: be \: (x2,y2) \\ let \: m = \frac{1}{2}$

$\frac{k - 0}{1 - 5} = \frac{1}{2}$

$\frac{k}{ - 4} = \frac{1}{2}$

$k = \frac{1}{2} \times - 4 = - 2$

Question 3,

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

$\frac{ - 4 - 2}{3 - x} = 2$

$\frac{ - 6}{3 - x } = 2$

$- 6 = 2(3 - x)$

$- 6 = 6 - 2x$

$- 6 - 6 = - 2x$

$- 2x = - 12$

$x = - 12 \div - 2 = 6$

Question 4,

$let \: (12, - 4) \: be \: (x1,y1) \\ let \: (r,2) \: be \: (x2,y2) \\ let \: m = - \frac{1}{2}$

$\frac{2 - ( - 4)}{r - 12} = - \frac{1}{2}$

$\frac{6}{r - 12} = - \frac{1}{2}$

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

$- r + 12 = 12$

$r = (12 - 12) \div - 1 = 0$