A line passes through (5 ,9 )(5,9) and (7 ,3 )(7,3). A second line passes through (3 ,1 )(3,1). What is one other point that the second line may pass through if it is parallel to the first line?
1 Answer
(4 ,-2)
Explanation:
The first step here is to calculate the gradient of the line passing through the 2 given points using the
color(blue)" gradient formula" 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)=(5,9)" and " (x_2,y_2)=(7,3)
rArr m = (3-9)/(7-5) = (-6)/2 = (-3)/1 = -3 Since the 2nd line is parallel then it's gradient equals -3
There are an infinite number of points that can lie on this line , but to find one we can use the definition of gradient and from (3 ,1) move 1 to the right and 3 down (This is equivalent to adding 1 to the x-coordinate and subtracting 3 from the y-coordinate )
hence (3 ,1) → (3+1 , 1-3) → (4 ,-2) is a point on the line.
Further points may be found in the same way.