y=f(x)=-1/2|x+6|-4y=f(x)=−12|x+6|−4 represents two lines y=-x/2-7y=−x2−7 and y=x/2-1y=x2−1 and solving them they intersect at (-6,-4)(−6,−4) (to solve just add them to get yy and then you get xx too).
Similarly y=g(x)=k|x+6|-10y=g(x)=k|x+6|−10 represents two lines y=kx+6k-10y=kx+6k−10 and y=-kx-6k-10y=−kx−6k−10 and solving them they intersect at (-6,-10)(−6,−10).
Assuming k=1k=1 we get the following graph. Observe that kite is formed with two points (-6,-4)(−6,−4) and (-6,-10)(−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/2-7y=−x2−7 and y=kx+6k-10y=kx+6k−10. Multiplying first by 2k2k we get 2ky=-kx-14k2ky=−kx−14k and adding this to second we get (2k+1)y=-8k-10(2k+1)y=−8k−10 and y=-(8k+10)/(2k+1)y=−8k+102k+1 and x=-2(-(8k+10)/(2k+1)+7)=(16k+20-28k-14)/(2k+1)=(-12k+6)/(2k+1)x=−2(−8k+102k+1+7)=16k+20−28k−142k+1=−12k+62k+1.
Hence, coordinates of third point are ((-12k+6)/(2k+1),-(8k+10)/(2k+1))(−12k+62k+1,−8k+102k+1) and area of kite is double the area of triangle formed by this point with (-6,-4)(−6,−4) and (-6,-10)(−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))|12∣∣∣(−6(−10+8k+102k+1)+6(−4+8k+102k+1)−−12k+62k+1(−4+10))∣∣∣
= 1/2|(60-(48k+60)/(2k+1)-24+(48k+60)/(2k+1)-(-72k+36)/(2k+1))|12∣∣∣(60−48k+602k+1−24+48k+602k+1−−72k+362k+1)∣∣∣
= 1/2|(36-(-72k+36)/(2k+1))|12∣∣∣(36−−72k+362k+1)∣∣∣
= 1/2|((72k+36+72k-36)/(2k+1))|12∣∣∣(72k+36+72k−362k+1)∣∣∣
= 1/2|(144k)/(2k+1)|12∣∣∣144k2k+1∣∣∣
And area of kite is (144k)/(2k+1)144k2k+1
As (144k)/(2k+1)=18144k2k+1=18
2k+1=144/18=82k+1=14418=8 and k=7/2k=72
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]}
