Question

6 The midpoint of the line segment joining

Answer 1

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 :

$m = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2} )$

Let (x1,y1) be coordinate A (h,4),

Let (x2,y2) be coordinate B (-5,k),

Let mid-point be (3,-2),

$(3 \: , - 2) = ( \frac{h - 5}{2} , \frac{4 + k}{2} )$

$by \: comparison, \:$

$\frac{h - 5}{2} = 3$

$h - 5 = 6$

$h = 11$

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

$4 + k = - 4$

$k = - 8$

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 :

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

Let (x1,y1) be (11,4),

Let (x2,y2) be (-5,-8),

$m = \frac{4 - ( - 8)}{11 - ( - 5)}$

$m = \frac{12}{16}$

$m = \frac{3}{4}$

The gradient of perpendicular line is opposite of line AB and when both gradient are multiplied, you should get -1 :

$m1 \times m2 = - 1$

Let m1 be the gradient of AB, m = 3/4,

Let m2 be the gradient of perpendicular line,

$\frac{3}{4} \times m2 = - 1$

$m2 = - \frac{ 4}{3}$

Last, we have to use the slope-form equation, y = mx + b and susbtitute the coordinates of M into the equation :

$y = mx + b$

Let m = -4/3,

Let x = 3,

Let y = -2,

$- 2 = - \frac{4}{3} (3) + b$

$b = 2$