Answer:
The first line:
y₁ = (2/3)*x + 6
Has the larger y-intercept, by 5 units.
Explanation:
Here we need to find the equation for each line.
First, some theory.
A linear relationship can be written as:
y = a*x + b
where a is the slope and y is the y-intercept.
We know that if the line passes through the points (x₁, y₁) and (x₂, y₂), then we can write the slope as:
a = (y₂ - y₁)/(x₂ -x₁)
And, if a line is:
y = a*x + b
a perpendicular line to that one must have a slope equal to:
-(1/a).
Now we can answer this question.
We know that the first line, let's call it y₁, passes through the points (3, 8) and (-3, 4), then its slope will be:
a = (8 - 4)/(3 - (-3)) = 4/6 = 2/3
then the line is something like:
y₁ = (2/3)*x + b
to find the value of b, we can use the fact that we know that the line passes through the point (3, 8)
this means that when x = 3, we must have y₁ = 8
replacing these in the above equation, we get:
8 = (2/3)*3 + b
8 = 2 + b
8 - 2 = b = 6
then the equation for this line is:
y₁ = (2/3)*x + 6
Now let's find the equation for the other line, that we will call y₂.
We know that this line is perpendicular to:
y = (1/3)*x - 2
The slope of that line is:
a = (1/3)
then the slope of a line perpendicular to that one will be:
slope = -(1/a) = -(1/1/3) = -3
slope = -3
then we have:
y₂ = -3*x + b
to find the value of b, we can use the fact that our line passes through the point (2, -5)
This means that when x = 2, we must have y₂ = -5
then:
-5 = -3*2 + b
-5 = -6 + b
-5 + 6 = b = 1
b = 1
then this equation is:
y₂ = -3*x + 1
Now we know both equations:
y₁ = (2/3)*x + 6
y₂ = -3*x + 1
Which equation does have the larger y-intercept?
We can see that the first line has an y-intercept of 6, and the second line has an y-intercept of 1, then the first line has the larger y-intercept, and is larger by 5 units.