Answer:
Question A, the values of h is 11 and k is -8 .
Question B, the equation is y = (-4/3)x + 2 .
Step-by-step explanation:
Question A, in order to find the value of h and k, you have to use the mid-point formula and do comparison :
![m = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2} )](https://tex.z-dn.net/?f=m%20%3D%20%28%20%5Cfrac%7Bx1%20%2B%20x2%7D%7B2%7D%20%20%2C%20%20%5Cfrac%7By1%20%2B%20y2%7D%7B2%7D%20%20%29)
Let (x1,y1) be coordinate A (h,4),
Let (x2,y2) be coordinate B (-5,k),
Let mid-point be (3,-2),
![(3 \: , - 2) = ( \frac{h - 5}{2} , \frac{4 + k}{2} )](https://tex.z-dn.net/?f=%283%20%5C%3A%20%2C%20-%202%29%20%3D%20%28%20%5Cfrac%7Bh%20-%205%7D%7B2%7D%20%2C%20%5Cfrac%7B4%20%2B%20k%7D%7B2%7D%20%29)
![by \: comparison, \:](https://tex.z-dn.net/?f=by%20%5C%3A%20comparison%2C%20%5C%3A%20)
![\frac{h - 5}{2} = 3](https://tex.z-dn.net/?f=%20%5Cfrac%7Bh%20-%205%7D%7B2%7D%20%20%3D%203)
![h - 5 = 6](https://tex.z-dn.net/?f=h%20%20-%205%20%3D%206)
![h = 11](https://tex.z-dn.net/?f=h%20%3D%2011)
![\frac{4 + k}{2} = - 2](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%20%2B%20k%7D%7B2%7D%20%20%3D%20%20-%202)
![4 + k = - 4](https://tex.z-dn.net/?f=4%20%2B%20k%20%3D%20%20-%204)
![k = - 8](https://tex.z-dn.net/?f=k%20%3D%20%20-%208)
Question B, given that line is perpendicular bisetor to AB means that the line touches mid-point which is M(3,-2). Using gradient formula :
![m = \frac{y2 - y1}{x2 - x1}](https://tex.z-dn.net/?f=m%20%3D%20%20%5Cfrac%7By2%20-%20y1%7D%7Bx2%20-%20x1%7D%20)
Let (x1,y1) be (11,4),
Let (x2,y2) be (-5,-8),
![m = \frac{4 - ( - 8)}{11 - ( - 5)}](https://tex.z-dn.net/?f=m%20%3D%20%20%5Cfrac%7B4%20-%20%28%20-%208%29%7D%7B11%20-%20%28%20-%205%29%7D%20)
![m = \frac{12}{16}](https://tex.z-dn.net/?f=m%20%3D%20%20%5Cfrac%7B12%7D%7B16%7D%20)
![m = \frac{3}{4}](https://tex.z-dn.net/?f=m%20%3D%20%20%5Cfrac%7B3%7D%7B4%7D%20)
The gradient of perpendicular line is opposite of line AB and when both gradient are multiplied, you should get -1 :
![m1 \times m2 = - 1](https://tex.z-dn.net/?f=m1%20%5Ctimes%20m2%20%3D%20%20-%201)
Let m1 be the gradient of AB, m = 3/4,
Let m2 be the gradient of perpendicular line,
![\frac{3}{4} \times m2 = - 1](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B4%7D%20%20%5Ctimes%20m2%20%3D%20%20-%201)
![m2 = - \frac{ 4}{3}](https://tex.z-dn.net/?f=m2%20%3D%20%20%20-%20%5Cfrac%7B%204%7D%7B3%7D%20)
Last, we have to use the slope-form equation, y = mx + b and susbtitute the coordinates of M into the equation :
![y = mx + b](https://tex.z-dn.net/?f=y%20%3D%20mx%20%2B%20b)
Let m = -4/3,
Let x = 3,
Let y = -2,
![- 2 = - \frac{4}{3} (3) + b](https://tex.z-dn.net/?f=%20-%202%20%3D%20%20%20-%20%5Cfrac%7B4%7D%7B3%7D%20%283%29%20%2B%20b)
![b = 2](https://tex.z-dn.net/?f=b%20%3D%202)