A line passes through #(4 ,1 )# and #(6 ,4 )#. A second line passes through #(3 ,5 )#. What is one other point that the second line may pass through if it is parallel to the first line?
1 Answer
Explanation:
The following result should be known.
#color(blue)"Parallel lines have equal slopes"# To calculate the slope use the
#color(blue)"gradient formula"#
#color(red)(bar(ul(|color(white)(2/2)color(black)(m=(y_2-y_1)/(x_2-x_1))color(white)(2/2)|)))#
where m represents the slope and# (x_1,y_1),(x_2,y_2)" 2 points on the line"# The 2 points here are (4 ,1) and (6 ,4)
let
# (x_1,y_1)=(4,1)" and " (x_2,y_2)=(6,4)#
#rArrm=(4-1)/(6-4)=3/2# Establish the equation of the line going through (3 ,5)
The equation of the line in
#color(blue)"point-slope form"# is.
#color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))#
where m is the slope and# (x_1,y_1)" a point on the line"#
#"Using "m=3/2" and " (x_1,y_1)=(3,5)#
#y-5=3/2(x-3)larrcolor(red)"in point-slope form"# distributing and simplifying gives an alternative version.
#y-5=3/2x-9/2#
#rArry=3/2x-9/2+5#
#rArry=3/2x+1/2larrcolor(red)" in slope-intercept form"# Selecting values for x and substituting into the equation will give corresponding values of y, and hence coordinate points.
#•x=1toy=3/2+1/2=2rArr(1,2)" is a point on the line"#
#•x=5toy=15/2+1/2=8rArr(5,8)" is a point on the line"#
graph{3/2x+1/2 [-10, 10, -5, 5]}