A line passes through #(5 ,8 )# and #(6 ,2 )#. A second line passes through #(1 ,3 )#. What is one other point that the second line may pass through if it is parallel to the first line?

There are 2 approaches we can take for this question.

(1) Using definition of gradient.

(2) Using equation of line passing through (1 ,3)

Either approach requires that we calculate the gradient of the line passing through (5 ,8) and (6 ,2).
We calculate the gradient using the #color(blue)"gradient formula"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(m=(y_2-y_1)/(x_2-x_1))color(white)(a/a)|)#
where #(x_1,y_1)" and " (x_2,y_2)" are 2 points"#

let #(x_1,y_1)=(5,8)" and " (x_2,y_2)=(6,2)#

#rArrm=(2-8)/(6-5)=(-6)/1=-6#

Method 1 Using the gradient

#color(blue)" Parallel lines have equal gradients"#
To get to another point on the line from (1 ,3) we move 6 down and 1 to the right (equivalent to subtracting 6 from y-coord and adding 1 to the x-coord)

Thus from (1 ,3) → (1+1 ,3-6) → (2 ,-3) is a point on line.

or we could move 1 left and 6 up (equivalent to subtracting 1 from x-coord and adding 6 to y-coord)

Thus from (1 ,3) → (1-1 ,3+6) → (0 ,9) is another point on the line.

Method 2 Using equation of line

Establish the equation in the form #color(red)(|bar(ul(color(white)(a/a)color(black)(y=mx+b)color(white)(a/a)|)))# where m is the gradient and b, the y-intercept.

We know m = -6 hence partial equation is y = -6x +b

Using (1 ,3) to find b → 3 = -6 + b → b = 9

#rArrcolor(red)(|bar(ul(color(white)(a/a)color(black)(y=-6x+9)color(white)(a/a)|)))# is the equation

Choosing any value for x and substituting into equation to find corresponding y-coordinate.

x = 0 : y = 0 + 9 = 9 → (0 ,9) is a point on line

x = 2 : y = -12 + 9 = -3 → (2 ,-3) is another point on the line

These are the same points generated using Method 1 above