The point of intersection of the lines \(\displaystyle\frac{x+1}{3}=\frac{y-3}{-1}=\frac{z-2}{2}\) and \(\displaystyle\frac{x-3}{1}=\frac{y-3}{-1}=\frac{z-8}{-1}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (5, 1, 6)
A general point of each line is
\(P(-1+3s,\ 3-s,\ 2+2s)\)
\(Q(3+t,\ 3-t,\ 8-t)\)
Comparing the y-coordinates:
\(3-s = 3-t\) gives \(s = t\)
Comparing the x-coordinates:
\(-1+3s = 3+s\)
\(s = 2,\quad t = 2\)
The z-coordinates agree too: \(2+2(2) = 6\) and \(8-2 = 6\). So the lines intersect at
\(P = (-1+6,\ 3-2,\ 2+4) = (5, 1, 6)\)
Why the other options are incorrect
Each of these points lies on only one of the lines:
Option (2) (2, 2, 4) — Incorrect. It is the point s = 1 of the first line; for the second line \(\frac{2-3}{1} = -1\) but \(\frac{2-3}{-1} = 1\).
Option (3) (4, 2, 7) — Incorrect. It is the point t = 1 of the second line; for the first line \(\frac{4+1}{3} = \frac{5}{3}\) but \(\frac{2-3}{-1} = 1\).
Option (4) (−1, 3, 2) — Incorrect. It is the given point of the first line; for the second line \(\frac{-1-3}{1} = -4\) but \(\frac{3-3}{-1} = 0\).