Consider $A(40\mathrm{\mid }220)A(40|220)$ and $B(100\mathrm{\mid }250)B(100|250)$.

Choose from $AA$: $x_1=40x\_1=40$ and $y_1=220y\_1=220$ and from $BB$: $x_2=100x\_2=100$ and $y_2=250y\_2=250$.

$m=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{y_2-y_1}{x_2-x_1}=\frac{250-220}{100-40}=\frac{30}{60}=0.5m=\backslash frac\{\backslash Delta\; y\}\{\backslash Delta\; x\}=\backslash frac\{y\_2-y\_1\}\{x\_2-x\_1\}=\backslash frac\{250-220\}\{100-40\}=\backslash frac\{30\}\{60\}=0.5$

Since the points $AA$-$DD$ all lie on a common line, it is sufficient to select only two points (for example $AA$ and $BB$). With their help you determine the slope of the line. To do this, subtract the $yy$-coordinate of point $AA$ from the $yy$-coordinate of point $BB$, and the $xx$-coordinate of point $AA$ from the $xx$-coordinate of point $BB$.

Then determine the $yy$-axis intercept by inserting a point on the straight line (for example C) into the straight line equation $y=mx+t\backslash \; \backslash \; y\; =\; mx\; +\; t\; \backslash$ and solving for $tt$.

Insert point $CC$ into the line equation $y=mx+ty\; =\; mx\; +\; t$ together with the previously calculated $m=0.5m\; =\; 0.5$:

With this we have determined both $mm$ and $tt$, so that our line equation is:

$y=0.5\cdot x+200\backslash \; \backslash \; y=0.5\backslash cdot\; x+200$

Insert any three $xx$-values into the line equation to get the respective $yy$-value, e.g. $x_1=0x\_1\; =\; 0$, $x_2=-100x\_2\; =\; -100$ and $x_3=300x\_3\; =\; 300$.

This gives us the following three points $DD$, $EE$ and $FF$:

$D(0\mathrm{\mid }200),E(-100\mathrm{\mid }150)D(0|200),\; E(-100|150)$ and $F(300\mathrm{\mid }350)F(300|350)$