Find the equation of the line joining the points (7,9,-3) and (11,-5,-3).?

Question

Find the equation of the line joining the points (7,9,-3) and (11,-5,-3).

Answer 1

This is a line in the x-y plane, with fixed z = -3 (by quick inspection)

So, we can find an equation relating x and y using point/slope form.

slope = (y2 - y1)/(x2 - x1) = (-5 - 9) / (11 - 7) = -14/4 = -7/2

Then, y - y1 = m(x - x1)
=> y + 5 = -7/2(x - 11)
=> y + 5 = -(7/2)x - 77/2
=> y = -(7/2)x - 67/2

Answer 2

Find the equation of the line joining the points P(7,9,-3) and
Q(11,-5,-3).

The directional vector v of the line is:

v = PQ = = <11, -5, -3> - <7, 9, -3> = <4, -14, 0>

Any non-zero multiple of v is also a directional vector of the line. Divide by 2.

v = <2, -7, 0>.

With the directional vector v and a point on the line we can write the equation of the line. Let's choose P(7,9,-3).

L = OP + tv
L = <7, 9, -3> + t<2, -7, 0>
where t is a scalar ranging over the real numbers