Refer to Exercise 11.3. Find the equations of the lines that pass through the points listed in Exercise 11.1.

Any x-y coordinate plane can be used to graph equations involving one or two variables. The following guidelines are valid in general: The coordinates of a point on the graph of an equation make the equation true, if a point's coordinates result in a true statement for an equation, the point is on the equation's graph.

Equation of straight line:

y= β₀+ β₁x .............(1)

If line passing through, (1, 1).

1= β₀+ β₁(1)

If line passing through, (5, 5).

5= β₀+ β₁(5)

Solve equation (2) & (3) simultaneously,

β₀+ β₁ -1 = β₀+ β₁(5)-5

β₁ -1 = 5β₁-5

4β₁= 4

β₁=1

Therefore, put β₁=1 in equation (2)

1= β₀+ 1(1)

1= β₀+ 1

β₀ =0

Now, put β₁=1 and β₀ =0 in equation (3)

y= β₀+ β₁x

y= 0+ 1x

y=x

Therefore, the required equation is y=x.

Equation of straight line:

y= β₀+ β₁x .............(1)

If line passing through, (0, 3).

3= β₀+ β₁(0)

β₀ = 3

If line passing through, (3, 0).

0= β₀+ β₁(3)

Therefore, put β₀= 3 in equation (2)

0= 3+ 3β₁

3β₁ = -3

β₁ = -1

Now, put β₁= -1 and β₀ = 3 in equation (3)

y= β₀+ β₁x

y= 3+ (-1)x

y= 3-x

Therefore, the required equation is y = 3-x.

Equation of straight line:

y= β₀+ β₁x .............(1)

If line passing through, (-1, 1).

1= β₀+ β₁(-1) .............(2)

If line passing through, (4, 2).

2= β₀+ β₁(4) ................(3)

Solve equation (2) & (3) simultaneously,

β₀+ β₁(-1) -1 = β₀+ β₁(4)-2

-β₁ -1 = 4β₁-2

4β₁+β₁ = 2-1

5β₁=1

β₁ =1/5

Therefore, put β₁= 1/5 in equation (2)

1= β₀+ (1/5)(-1)

1= β₀(-1/5)

β₀ = 1+(1/5)

β₀=(5+1)/5

β₀=6/5

Now, put β₁=(1/5) and β₀ =6/5 in equation (3)

y= β₀+ β₁x

y= (6/5)+ (1/5)x

y= (6/5)+(x/5)

Therefore, the required equation is y= (6/5)+(x/5).