a) The procedure to draw the straight line "2x-3y\\ge6" .
We draw the straight line
"2x-3y =6" ....(1)
by reducing it into "y=mx+c"
Equation (1) can be written as "3y=2x-6"
Dividing with '3' we get
"y= \\frac{2}{3}x-2""\\Rightarrow" slope "m= \\frac{2}{3}"
and y intercept "c= -2"
Since slope "m= \\frac{rise}{run}" , for a rise by 2 units, run will be 3
"\\Rightarrow" "(3,0)" is a point on the straight line.
Since y intercept "c=-2" , "(0,-2)" is a point on the straight line.
Joining these two points and extending it we get a straight line "2x-3y=6" .
Since "2x-3y\\ge6" is satisfied by the points "x\\ge3" and "y\\le-2"
The equation is the points satisfying "x\\ge3" and "y\\le-2."
c) Substituting "x=0" in the straight line we get "2(0)-3y=6"
"\\Rightarrow y=-2"
"(0,-2)" is a point of intersection with y axis.
b) The point of intersection with "x" axis is given by substituting "y=0" in the straight line.
"\\Rightarrow 2x-3(0)=6"
"\\Rightarrow" "x=3"
"\\Rightarrow" "(3,0)" is the point of intersection with "x" axis.
Therefore, the points of intersection with axes are "(3,0)" and "(0,-2)."
d) Since the straight line cuts the x axis at "x=3" and y axis at "y=-2" and "2x-3y\\ge6" , the origin "(0,0)" lies to the left of the straight line.
e) Since the straight line includes "equality", the points satisfying "2x-3y=6" are also to be graphed.
Hence it is a solid line.
f) Right side of the line to be shaded.
g) Since "2x-3y\\ge6" satisfied by "x\\ge3" and "y\\le-2" .
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