Question

The co-ordinates of the point of intersection of the lines $\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z-1}{1}$ and $\frac{x}{2}=\frac{z+3}{3},y=-1$ are:

• A. $(2,1,0)$
• B. $(-2,1,0)$
• C. $(2,-1,0)$
• D. $(-2,-1,0)$

Answer 1

Answer: (C)

$(2,-1,0)$

Let $\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z-1}{1}=k$, for a real $k$

Thus, any point on the line may be represented as:

$(k+3,-k-2,k+1)$

If the point also lies on the second line

$\frac{x}{2}=\frac{z+3}{3},y=-1$,

Then the following must hold:

$\frac{k+3}{2}=\frac{k+4}{3},-k-2=-1$

It follows that

$k=-1$

Hence, the point common to both lines is given by:

$(-1+3,1-2,-1+1)$

i.e

$(2,-1,0)$