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Graphically, solve the following pair of equations:
$2x+y=6$
$2x-y+2=0$
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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