#y=f(x)=-1/2|x+6|-4# represents two lines #y=-x/2-7# and #y=x/2-1# and solving them they intersect at #(-6,-4)# (to solve just add them to get #y# and then you get #x# too).
Similarly #y=g(x)=k|x+6|-10# represents two lines #y=kx+6k-10# and #y=-kx-6k-10# and solving them they intersect at #(-6,-10)#.
Assuming #k=1# we get the following graph. Observe that kite is formed with two points #(-6,-4)# and #(-6,-10)# vertically aligned and two other points formed by positively and negatively sloping pair of lines.
graph{(y+x/2+7)(y-x/2+1)(y+x+16)(y-x+4)=0 [-16.54, 3.46, -11.68, -1.68]}
Hence, let us consider intersection of #y=-x/2-7# and #y=kx+6k-10#. Multiplying first by #2k# we get #2ky=-kx-14k# and adding this to second we get #(2k+1)y=-8k-10# and #y=-(8k+10)/(2k+1)# and #x=-2(-(8k+10)/(2k+1)+7)=(16k+20-28k-14)/(2k+1)=(-12k+6)/(2k+1)#.
Hence, coordinates of third point are #((-12k+6)/(2k+1),-(8k+10)/(2k+1))# and area of kite is double the area of triangle formed by this point with #(-6,-4)# and #(-6,-10)#.
Area of triangle is
#1/2|(-6(-10+(8k+10)/(2k+1))+6(-4+(8k+10)/(2k+1))-(-12k+6)/(2k+1)(-4+10))|#
= #1/2|(60-(48k+60)/(2k+1)-24+(48k+60)/(2k+1)-(-72k+36)/(2k+1))|#
= #1/2|(36-(-72k+36)/(2k+1))|#
= #1/2|((72k+36+72k-36)/(2k+1))|#
= #1/2|(144k)/(2k+1)|#
And area of kite is #(144k)/(2k+1)#
As #(144k)/(2k+1)=18#
#2k+1=144/18=8# and #k=7/2#
and kite appears as
graph{(y+x/2+7)(y-x/2+1)(y+7/2x+31)(y-7/2x-11)=0 [-16.5, 3.5, -12.2, -2.2]}
