draw the graph representing the equation x - y= 1 and 2 X + 3 Y = 12 on the same graph paper find the area of a triangle formed by these lines at x-axis and the y-axis
Answers
Step-by-step explanation:
Graph of the equation 2x + 3y - 12 = 0
we have,
2x + 3y = 12 =2x+3y=12=
2x = 12 - 3y =
→ `x = (12- 3y )/ 2°
Putting y = 4, we get `x ( 12 - 3 xx 4)/2 = 0
Putting y = 2, we get `x = (12 - 3 xx = =4)/2= 0
Thus, we have the following table for the p table for the points on the line 2x + 3y = 12
⠀⠀⠀⠀⠀⠀⠀\{\fcolorbox{red}{pink}{x}{
x
\{\fcolorbox{red}{pink}{0}{
0
\{\fcolorbox{yellow}{green}{3}{
3
⠀⠀⠀⠀⠀⠀⠀\{\fcolorbox{green}{red}{y}{
y
\{\fbox{4}{
4
\{\fbox{2}{
2
Plotting points A ( 0,4 ) , B (3,2) on the graph paper and drawing a line
passing through them
we obtain graph of the equation.
Graph of the equation
Graph of the equation x-y - 1 :
We have x −− == y = 1 = x = 1 ++ y
Thus, we have the following table for the points the line x - y = 1
⠀⠀⠀⠀⠀⠀⠀\{\fcolorbox{red}{pink}{x}{
x
\{\fcolorbox{red}{pink}{1}{
1
\{\fcolorbox{yellow}{green}{0}{
0
⠀⠀⠀⠀⠀⠀⠀\{\fcolorbox{green}{red}{y}{
y
\{\fbox{0}{
0
\{\fbox{-1}{
-1
Plotting points C(1,0) and D (0,-1) on the same graph paper drawing a line passing through
the m, we obtain the graph of the line represents by the equation x - y = 1
Clearly two lines intersect at A (3, 2)
The graph of time 2x + 3y = 12 = intersect with y-axis at B (0,4) and the graph of the line
X - y = 1 intersect with y axis at c ( =
0,-1).
So, the vertices of the triangle formed by thee two straight lines and y-axis are A (3,2) and
B( 0.4) and C (0,-1)
Now,
Area of `AABC = 1/2` [ Base xx =
Height]
= 1/2 ( Bc xx AB)` =
= 1/2 ( 5 + 3)`
= 15/2 sq .units