Plot the graph of the following linear equation 2(x+3)-3(y+1)=0. Also the following equation
a) Writ the quadrant in which the line segment intercepted between the axes lie.
b) shade the region formed by the line and the axes.
c) Write the vertices of the triangle so formed.

The given equation is,   2x+3 - 3y+1 = 02x+6-3y-3 = 02x-3y+3 = 02x + 3 = 3yy = 2x+33When x = 0; y = 1When x = -1.5, y = 0When x = 3, y = 3

In tabular form,
 
x 0 -1.5 3
y 1 0 3


GRAPH :



1. The line segment intercepted between the axes lies in the second quadrant.2.The  region between line and the coordinate axes is shaded as shown in the above figure.3. The triangle formed is AOB. The coordinates of the vertices are :        A0,1; O0,0;B-1.5,0

 

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