Answer:
The given point satisfies both equations of a line, hence is common to the two lines.
Explanation:
Given two lines, one joining the points (6,-7,0) to (16,-19,-4), and the other joining the points (0,3,-6) to (2,-5,10), you want to show that the point (1, -1, 2) is common to those lines.
Equation of a line
The equation of a line through points (x1, y1, z1) and (x2, y2, z2) can be written:
(x -x1)/(x2 -x1) = (y -y1)/(y2 -y1) = (z -z1)/(z2 -z1)
For the first pair of points, the equation of the line is ...
(x -6)/(16 -6) = (y +7)/(-19 +7) = (z -0)/(-4 -0)
(x -6)/10 = -(y +7)/12 = -z/4 . . . . . line 1
For the second pair of points, the equation of the line is ...
(x -0)/(2 -0) = (y -3)/(-5 -3) = (z +6)/(10 +6)
x/2 = -(y -3)/8 = (z +6)/16 . . . . . line 2
Point on the line?
If the given point is on each of these lines, then it will satisfy each of the equations.
Line 1: (1 -6)/10 = -(-1 +7)/12 = -2/4 ⇒ -5/10 = -6/12 = -2/4 . . . true
The given point is on Line 1.
Line 2: (1)/2 = -(-1-3)/8 = (2+6)/16 ⇒ 1/2 = 4/8 = 8/16 . . . true
The given point is on Line 2.
The given point is common to both lines.
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Alternate solution
For points A and B and parameter p, we can write the parameterized equation of each line as ...
(x, y, z) = A +p(B -A)
Using parameters t and s for the two lines, we get ...
Line 1: (x, y, z) = (6, -7, 0) +t(16 -6, -19 +7, -4 -0) = (6 +10t, -7 -12t, 0 -4t)
Line 2: (x, y, z) = (0, 3, -6) +s(2 -0, -5 -3, 10 +6) = (0 +2s, 3 -8s, -6 +16s)
A common point will be found where (x, y, z) for some value of t matches (x, y, z) for some value of s. We can find those values of s and t by solving simultaneous equations for any pair of the variables (x, y, z). The equations are ...
- 6 +10t = 2s
- -7 -12t = 3 -8s
- -4t = -6 +16s
By any of your favorite methods, these equations have the solution (t, s) = (-0.5, 0.5). The point the lines have in common is ...
(6 +10(-0.5), -7 -12(-0.5), 0 -4(-0.5)) = (6 -5, -7 +6, 2) = (1, -1, 2)
The common point to the two lines is (1, -1, 2).