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# Graphically, solve the following pair of equations:

$ 2 x+y=6 $

$ 2 x-y+2=0 $

Find the ratio of the areas of the two triangles formed by the lines representing these equations with the $ x $-axis and the lines with the $ y $-axis.

Given:

The given equations are:

$2x+y=6$

$2x-y+2=0$

To do:

We have to solve the given system of equations and find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.

Solution:

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

For equation $2x+y-6=0$,

$y=6-2x$

If $x=0$ then $y=6-2(0)=6-0=6$

If $x=3$ then $y=6-2(3)=6-6=0$

$x$ | $0$ | $3$ |

$y$ | $6$ | $0$ |

For equation $2x-y+2=0$,

$y=2x+2$

If $x=0$ then $y=2(0)+2=2$

If $x=-1$ then $y=2(-1)+2=-2+2=0$

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

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

The equation of y-axis is $x=0$.

The equation of x-axis is $y=0$.

The above situation can be plotted graphically as below:

The lines AB, CD, BD and AC represent the equations $2x+y=6$, $2x-y+2=0$, x-axis and y-axis respectively.

The points of intersection of the lines AB, CD and x-axis taken in pairs are the vertices of the triangle formed with x-axis.

Hence, the vertices of the triangle formed with x-axis are $(1,4), (-1,0)$ and $(3,0)$.

We know that,

Area of a triangle$=\frac{1}{2}bh$

In the graph, the height of the triangle is the distance between point E and BD.

Height of the triangle$=4$ units.

Base of the triangle$=$Distance between the points B and D.Base of the triangle$=1+3=4$ units.

Area of the triangle formed by the given lines and x-axis$=\frac{1}{2}\times4\times4$

$=8$ sq. units.

The points of intersection of the lines AB, CD and y-axis taken in pairs are the vertices of the triangle formed with y-axis.

Hence, the vertices of the triangle formed with y-axis are $(0,6), (0,2)$ and $(1,4)$.

We know that,

Area of a triangle$=\frac{1}{2}bh$

In the graph, the height of the triangle is the distance between point E and AC.

Height of the triangle$=1$ units.

Base of the triangle$=$Distance between the points A and C.Base of the triangle$=6-2=4$ units.

Area of the triangle formed by the given lines and y-axis$=\frac{1}{2}\times1\times4$

$=2$ sq. units.

The ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis is $8:2=4:1$.

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