## Elementary Geometry for College Students (7th Edition) Clone

(a) The point $(2,1,-3)$ lies on the line. (b) The point $(8,5,-11)$ lies on the line.
$(x,y,z) = (2,1,-3)+n(3,2,-4)$ (a) If $(2,1,-3)$ lies on the line, then there is a real number $n$ such that $(2,1,-3)+n(3,2,-4) = (2,1,-3)$ $x$: If $2+n(3) = 2$, then $n = 0$ $y$: If $1+n(2) = 1$, then $n = 0$ $z$: If $-3+n(-4) = -3$, then $n = 0$ Since the required value of $n$ is the same for $x,y,$ and $z$, the point $(2,1,-3)$ lies on the line. (b) If $(8,5,-11)$ lies on the line, then there is a real number $n$ such that $(2,1,-3)+n(3,2,-4) = (8,5,-11)$ $x$: If $2+n(3) = 8$, then $n = 2$ $y$: If $1+n(2) = 5$, then $n = 2$ $z$: If $-3+n(-4) = -11$, then $n = 2$ Since the required value of $n$ is the same for $x,y,$ and $z$, the point $(8,5,-11)$ lies on the line.