A line passes through #(6 ,5 )# and #(8 ,3 )#. A second line passes through #(0 ,1 )#. What is one other point that the second line may pass through if it is parallel to the first line?

1 Answer
Aug 2, 2016

Answer:

The equation of the second line is: #" "y=-x+1#

Explanation:

If the two lines are parallel then they have the same gradient (slope).

For the first line:
Let point 1 be #P_1->(x_1,y_1) = (6,5)#
Let point 2 be #P_2->(x_2,y_2)=(8,3)#

For the second line:
Let point 3 be #P_3->(x_3,y_3) =(0,1)#

For both lines:
Let the gradient (slope) be #m#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider the first line with "P_1 and P_2)#

gradient #=m=("change in the y axis")/("change in the x axis")#

#=>m=(y_2-y_1)/(x_2-x_1) =(3-5)/(8-6) = (-2)/(2) = -1#

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider the second line with "P_3)#

#color(brown)("as this line is parallel to the first it has the same gradient (m)")#

The standard form equation for a straight line is:

#y=mx+c#

But #m=-1# giving

#y=(-1)x+c" "->" "y=-x+c#

We know that this line passes through the point #P_3#

So the value of the coordinates for #P_3# must be true for this equation giving:

#P_3->(x_3,y_3)->(color(magenta)(0),color(green)(1))" " =>" "color(green)( 1)" "=" "-1(color(red)(0))+c#

Thus #c=1# giving

#color(brown)(y=mx+c)color(blue)(" "->" "y=-x+1#