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# Determine, graphically, the vertices of the triangle formed by the lines

$ y=x, 3 y=x, x+y=8 $

Given:

The equations of the sides of the given triangle are:

$y=x, 3y=x$ and $x+y=8$

To do:

We have to determine the vertices of the given triangle.

Solution:

To represent the above equations graphically we need at least two solutions for each of the equations.

For equation $y=x$,

If $x=0$ then $y=0$

If $x=1$ then $y=1$

$x$ | $0$ | $1$ |

$y$ | $0$ | $1$ |

For equation $3y=x$,

$y=\frac{x}{3}$

If $x=0$ then $y=\frac{0}{3}=0$

If $x=6$ then $y=\frac{6}{3}=2$

$x$ | $0$ | $6$ |

$y$ | $0$ | $2$ |

For equation $x+y=8$,

$y=8-x$

If $x=5$ then $y=8-5=3$

If $x=6$ then $y=8-6=2$

$x$ | $5$ | $6$ |

$y$ | $3$ | $2$ |

The above situation can be plotted graphically as below:

The lines AB, AC and CD represent the equations $y=x$, $y=3x$ and $x+y=8$ respectively.

As we can see, the points of intersection of the lines AB, AC and CD taken in pairs are the vertices of the given triangle.

Hence, the vertices of the given triangle are $(0,0), (6,2)$ and $(4,4)$.

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