Question

Which equation represents graph (1)
(1)

(2)

(3)

(4)

A. \[z = y + 1\]
B. \[y = z + 1\]
C. \[z + y = 1\]
D. \[y = x - 1\]

Answer 1

Hint: We will write the solutions of graph (1) and then we will check among these four equations which equation is satisfied by these solutions. The equation that will satisfy these two solutions is the equation representing the graph (1).

Complete step-by-step answer:
From the figure (1), we can clearly see the points of the graph.
Hence, the points are \[\left( {1,0} \right)\] and \[\left( {0,1} \right)\].
Now, we will check that these points satisfy which equation. The equation which satisfy these points is the equation representing graph (1)
Consider the equation \[z = y + 1\] …. (1)
Now taking the first point i.e. \[\left( {1,0} \right)\]
Putting \[y = 1,z = 0\] in equation (1), we get
 \[ \Rightarrow 0 = 1 + 1\]
On simplification we get,
 \[ \Rightarrow 0 = 2\]
Which is not true. Hence, the equation \[z = y + 1\] does not represent graph (1).
Consider the equation \[y = z + 1\] …. (2)
Now taking the first point i.e. \[\left( {1,0} \right)\]
Putting \[y = 1,z = 0\] in the equation (2), we get
 \[ \Rightarrow 0 + 1 = 1\]
On simplification we get,
 \[ \Rightarrow 1 = 1\]
Which is true. Now, we will check for another point \[\left( {0,1} \right)\].
 \[ \Rightarrow 0 = 1 + 1\]
On simplification we get,
 \[ \Rightarrow 0 = 2\]
Which is not true. That means \[\left( {0,1} \right)\] does not satisfy the equation \[y = z + 1\].
Hence, the equation \[y = z + 1\] does not represent graph (1).
 i.e. \[\left( {1,0} \right)\] satisfies the equation \[z + y = 1\].
Consider the equation\[z = y + 1\]. … (3)
Now taking the first point i.e. \[\left( {1,0} \right)\]
Putting \[y = 1,z = 0\] in the equation (3), we get
 \[ \Rightarrow 0 + 1 = 1\]
On simplification we get,
 \[ \Rightarrow 1 = 1\]
 i.e. \[\left( {1,0} \right)\] satisfies the equation \[z + y = 1\].
Now we take the second point i.e. \[\left( {0,1} \right)\]
Putting \[y = 0,z = 1\] in the equation (3), we get
 \[ \Rightarrow 0 + 1 = 1\]
On simplification we get,
 \[ \Rightarrow 1 = 1\]
Which is true.
 Which means the given points are satisfied by this equation. Hence, \[z + y = 1\] is the equation representing graph (1).
Hence, option (C) is the correct option.

Note: A linear equation in two variables has infinitely many solutions. The graph of every linear equation in two variables is a straight line. Each point on the graph of the linear equation is a solution of the linear equation.