Theequationoflinepis y = -4 5 x + 4 . Line q , whichisparalleltoline p , includesthepoint (4, -3) . What istheequationofline q ?

Question

Equations of parallel and perpendicular lines

 

Answer 1

The equation of the line P is y= -4.5 x + 4 

The equation of the line Q is y= -4.5 x + c  [since it is parallel to the line P]

The lile passes through the point (4, -3),

Thus, -3 = -4.5X 4 +c

thus, c= 4.5x 4 -3 = 18-3= 15

thus the equation of the line Q is y=-4.5 x +15

Answer 2

y = -45x + 4

 

Since line q is parallel, it has the same slope, which is -45, by inspection.  Then the equation for line q is

 

y = -45x + b.  We find b from the given point (4,-3);

 

-3 = -45(4) + b ⇒ b = 177

 

So the equation of line q is

 

y = -45x + 177

Answer 3

p and q are parallel, so q is also of the form y=-45x+h. To find h, we use the point (4,-3). This point is on q so we must have -3=(-45)(4)+h. Hence h=177 and so the equation of the line q is y=-45x+177.