Question

Complete the equations of the system in slope-intercept form. Use a decimal for the slope if necessary.

Line 1
x y
0 3
1 5
Line 2
x y
–1 1
–2 –1

Line 1: y = x +
Line 2: y = x +

Answer 1

@MAli13chineta

Answer 2

im not sure sorry @PlanSlam @wio @TheSmartOne

Answer 3

Okay, so the thing to understand about lines, is that the change in \(y\) and change in \(x\) are always proportional.

Answer 4

ok

Answer 5

For line 1, the change in \(y\) is given by \(y_2-y_1=5-3=2\).
The change in \(x\) is given by \(x_2-x_1=1-0=1\).
You get the slope by dividing change in \(y\) by change in \(x\).

Answer 6

For line 1, the slope is \(m=2/1=2\)

Answer 7

so y=2 x+2?

Answer 8

The \(y\) intercept is what you get when you let \(x=0\).

Answer 9

Line 1 already tells us that when \(x=0\), we have \(y=3\)

Answer 10

So with a slope of \(2\) and \(y\) intercept of \(3\), our line is:\[
y=2x+3
\]

Answer 11

In general, if you have two points \((x_1,y_1)\) and \((x_2,y_2)\), you can find the equation for the line by solving for \(y\): \[
\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}
\]So for line 2, we would start with:\[
\frac{y-(1)}{x-(-1)}=\frac{(-1)-(1)}{(-2)-(-1)}
\]

Answer 12

Using algebra on:\[
\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}
\]to solve for \(y\) would give:\[
y = \frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1
\]

Answer 13

Sorry @wio my connection went fuzzy and had to refresh