Answer: lines j and k are perpendicular
Explanation:
Let's find the equations of lines j and k
![\displaystyle\\\boxed{(x-x_1)/(x_2-x_1)=(y-y_1)/(y_2-y_1) }](https://img.qammunity.org/2023/formulas/mathematics/high-school/nvr35m08htwkwky46n85st7z9vkyfzvntc.png)
Line j: (8,2) (-2,-2)
x₁= 8 x₂=-2 y₁=2 y₂=-2
![\displaystyle\\(x-8)/(-2-8)=(y-2)/(-2-2) \\\\(x-8)/(-10)=(y-2)/(-4) \\\\](https://img.qammunity.org/2023/formulas/mathematics/high-school/4t3v0rf0r94srq4ce1p769bq43l9uu9l9y.png)
Multiply both parts of the equation by -4:
![\displaystyle\\(2)/(5) (x-8)=y-2\\\\(2)/(5) x-(16)/(5) =y-2\\\\(2)/(5) x-3.2+2=y-2+2\\\\(2)/(5)x-1.2=y\\\\ Thus,\ y=(2)/(5)x-1.2](https://img.qammunity.org/2023/formulas/mathematics/high-school/jdr3ohya4yzzotlrff2e11f96mr2vlxg0e.png)
Line k: (-4,3) (-6,8)
x₁=-4 x₂=-6 y₁=3 y₂=8
![\displaystyle\\(x-(-4))/(-6-(-4)) =(y-3)/(8-3) \\\\(x+4)/(-6+4)=(y-3)/(5) \\\\(x+4)/(-2) =(y-3)/(5)](https://img.qammunity.org/2023/formulas/mathematics/high-school/minxsq6po3k3earu54pvuih5emqojwkoqc.png)
Multiply both parts of the equation by 5:
![\displaystyle\\-(5)/(2)(x+4)=y-3\\\\ -(5)/(2)x-10=y-3\\\\ -(5)/(2)x-10+3=y-3+3\\ -(5)/(2)x-7=y\\Thus,\ y=-(5)/(2) x-7](https://img.qammunity.org/2023/formulas/mathematics/high-school/1zd5awfujrjg6sk84qndhzerbasvnr5i1j.png)
Hence, lines j and k are perpendicular