The normal vector to the plane <em>x</em> + 3<em>y</em> + <em>z</em> = 5 is <em>n</em> = (1, 3, 1). The line we want is parallel to this normal vector.
Scale this normal vector by any real number <em>t</em> to get the equation of the line through the point (1, 3, 1) and the origin, then translate it by the vector (1, 0, 6) to get the equation of the line we want:
(1, 0, 6) + (1, 3, 1)<em>t</em> = (1 + <em>t</em>, 3<em>t</em>, 6 + <em>t</em>)
This is the vector equation; getting the parametric form is just a matter of delineating
<em>x</em>(<em>t</em>) = 1 + <em>t</em>
<em>y</em>(<em>t</em>) = 3<em>t</em>
<em>z</em>(<em>t</em>) = 6 + <em>t</em>
Find x you mean?
substract both side by 2x
5x=-35
divide both side by 5
x=-7
->D
Answer:
C (1,11) and (-1,-5)
Step-by-step explanation:
Answer:
6/5
Step-by-step explanation:
We can find the slope given two points by
m = (y2-y1)/(x2-x1)
= (1 - -5)/(7-2)
= (1+5)/(7-2)
=6/5