A) AB and CB is perpendicular. Thus, the gradient of AB multiplied by the gradient of CB is -1 (m1m2=-1) .
-1/2(m2) = -1
m2 = 2
CB : y = 2x + c
c is the y-intercept. It is given in the diagram, on the y-axis - the point B.
CB : y = 2x +2
Now, substitute the coordinate C into the equation.
k = 2(4) + 2
k = 10
b) In order to find the are, we will use a formula that require all coordinates of the points to be revealed.
Find coordinate A.
A(p,10). Why 10? --> A and C is located at the same line that is parallel to the x-axis. They have the same y-value/height.
AC : y = 10, AC is parallel to the x-axis.
AB : y = -x/2 + 2
Substitute the y value into equation AB to find p.
10 = -x/2 + 2
x = -16
Use the formula to find the area of a polygon. It is given in Form 4, Chapter 6.
=1/2 | (-16)(10)+(4)(2)+(0)(10) - (4)(10)+(0)(10)+(-16)(2) |
c) Lets call our new line DE.
DE : y = mx + c
It is parallel to AB. That means both have the same gradient.
DE : y = -x/2 + c OR y = (-1/2)(x) + c
Now, lets find the midpoint of BC.
=( (4+0)/2 , (10+2)/2 )
Substitute the coordinate of the midpoint BC into the equation to find c, y-intercept.
6 = -2/2 + c
c = 7
Rewrite the equation.
y = -x/2 + 7