Exercises: Linear functions as lines in coordinate systems

Straight line AB:

$A=(-2|1)$ and $B=(6|1)$

Determine the slope m with the difference quotient.

$m=\frac{1-1}{-2-6}=\frac{0}{-8}=0$

So the straight line is parallel to the x-axis.The y-intercept t is therefore equal to the y-coordinate of the two points.

$t=1$

Put together the equation of the line.

$y=mx+t=0\cdot x+1=1$

Straight line AC:

$A=(-2|1)$ and $C=(4|5)$

Determine the slope m with the difference quotient.

$m=\frac{1-5}{-2-4}=\frac{-4}{-6}=\frac{2}{3}$

So the straight line is parallel to the x-axis.The y-intercept t is therefore equal to the y-coordinate of the two points.

$y=mx+t$

Put a point for example. A .

$1=\frac{2}{3}(-2)+t$

Solve for t .

$t=1-\frac{-4}{3}=1+\frac{4}{3}=\frac{7}{3}$

Put together the equation of the line.

$y=mx+t=\frac{2}{3}x+\frac{7}{3}$

Straight line BC:

$B=(6|1)$ and $C=(4|5)$

Determine the slope m with the difference quotient.

$m=\frac{1-5}{6-4}=\frac{-4}{2}=-2$

Set up the equation for the y-intercept.

$y=mx+t$

Put a point, for example B .

$1=-2\cdot 6+t$

Solve for t .

$t=1-(-2)\cdot 6=1+12=13$

Put together the equation of the line.

$y=mx+t=-2x+13$

The perimeter of the triangle is calculated from the lengths of the segments between the points. Use the well-known formula.

$d(P_1,P_2)=\sqrt{(\mathrm{\Delta }x{)}^2+(\mathrm{\Delta }y{)}^2}$

Enter two points for each route.

Side AB:

$\mathrm{\Delta }x=-2-6=-8$

$\mathrm{\Delta }y=1-1=0$

Plug into the formula. Note that only positive results are relevant for lengths.

$d(A,B)=\sqrt{(-8{)}^2+0^2}=\sqrt{64}=8$

So this side has length 8.

Side AC:

$\mathrm{\Delta }x=-2-4=-6$

$\mathrm{\Delta }y=1-5=-4$

Plug into the formula. Note that only positive results are relevant for lengths.

$d(A,C)=\sqrt{(-6{)}^2+(-4{)}^2}=\sqrt{36+16}$ $=\sqrt{52}\approx 7a21$

So this side is about length 7.2.

Side BC:

$\mathrm{\Delta }x=6-4=2$

$\mathrm{\Delta }y=1-5=-4$

Plug into the formula. Note that only positive results are relevant for lengths.

$d(A,C)=\sqrt{2^2+(-4{)}^2}=\sqrt{4+16}=\sqrt{20}\approx 4a47$

So this side is about length 4.5.

$\Rightarrow$ The perimeter is equal to 8 + 7.2 + 4.5 = 19.7.

$A=\frac{1}{2}\cdot \text{Hight}\cdot base$

Choose AB as the base because it is parallel to the x-axis. This makes the height parallel to the y-axis.

Height h is equal to the distance from C to AB.

Since the height is parallel to the y-axis, the distance is calculated simply by comparing the y-coordinates.

The distance of point C from the baseline is $h=5-1=4$

Inserted into the formula, this gives: $A=\frac{1}{2}\cdot 4\cdot 8=16$