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 :
[tex]m = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2} )[/tex]
Let (x1,y1) be coordinate A (h,4),
Let (x2,y2) be coordinate B (-5,k),
Let mid-point be (3,-2),
[tex](3 \: , - 2) = ( \frac{h - 5}{2} , \frac{4 + k}{2} )[/tex]
[tex]by \: comparison, \: [/tex]
[tex] \frac{h - 5}{2} = 3[/tex]
[tex]h - 5 = 6[/tex]
[tex]h = 11[/tex]
[tex] \frac{4 + k}{2} = - 2[/tex]
[tex]4 + k = - 4[/tex]
[tex]k = - 8[/tex]
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 :
[tex]m = \frac{y2 - y1}{x2 - x1} [/tex]
Let (x1,y1) be (11,4),
Let (x2,y2) be (-5,-8),
[tex]m = \frac{4 - ( - 8)}{11 - ( - 5)} [/tex]
[tex]m = \frac{12}{16} [/tex]
[tex]m = \frac{3}{4} [/tex]
The gradient of perpendicular line is opposite of line AB and when both gradient are multiplied, you should get -1 :
[tex]m1 \times m2 = - 1[/tex]
Let m1 be the gradient of AB, m = 3/4,
Let m2 be the gradient of perpendicular line,
[tex] \frac{3}{4} \times m2 = - 1[/tex]
[tex]m2 = - \frac{ 4}{3} [/tex]
Last, we have to use the slope-form equation, y = mx + b and susbtitute the coordinates of M into the equation :
[tex]y = mx + b[/tex]
Let m = -4/3,
Let x = 3,
Let y = -2,
[tex] - 2 = - \frac{4}{3} (3) + b[/tex]
[tex]b = 2[/tex]
