Math, asked by jarvickanishk619, 8 months ago

The graph of which of the following linear equations will pass through origin a) x + y =4 b) x – y= 2 c) x= 3y d) 3 = 2x + y

Answers

Answered by AlluringNightingale
24

Answer :

c). x = 3y

Solution :

★ Let's check whether x + y = 4 passes through origin (0,0) or not .

Putting x = 0 and y = 0 , we get ;

=> LHS = x + y

=> LHS = 0 + 0

=> LHS = 0

RHS = 4

Clearly ,

LHS ≠ RHS , thus the equation x + y = 4 doesn't passes through origin (0,0) .

××××××××××××××××××××××××××××××××××××

★ Let's check whether x - y = 2 passes through origin (0,0) or not .

Putting x = 0 and y = 0 , we get ;

=> LHS = x - y

=> LHS = 0 - 0

=> LHS = 0

RHS = 2

Clearly ,

LHS ≠ RHS , thus the equation x - y = 2 doesn't passes through origin (0,0) .

××××××××××××××××××××××××××××××××××××

★ Let's check whether x = 3y passes through origin (0,0) or not .

Putting x = 0 and y = 0 , we get ;

=> LHS = x

=> LHS = 0

=> RHS = 3y

=> RHS = 3×0

=> RHS = 0

Clearly ,

LHS = RHS , thus the equation x = 3y passes through origin (0,0) .

××××××××××××××××××××××××××××××××××××

★ Let's check whether 3 = 2x + y passes through origin (0,0) or not .

Putting x = 0 and y = 0 , we get ;

LHS = 3

=> RHS = 2x + y

=> RHS = 2×0 + 0

=> RHS = 0 + 0

=> RHS = 0

Clearly ,

LHS ≠ RHS , thus the equation 3 = 2x + y doesn't passes through origin (0,0) .

Hence ,

The linear equation x = 3y passes through origin.

××××××××××××××××××××××××××××××××××××

Note :

If an equation doesn't contain any constant term , then it always passes through origin .

The linear equation x = 3y doesn't contain any constant term , thus it passes through origin .

Answered by krishna2098
10

Answer:

Step-by-step explanation:

Let's check whether x + y = 4 passes through origin (0,0) or not .

Putting x = 0 and y = 0 , we get ;

=> LHS = x + y

=> LHS = 0 + 0

=> LHS = 0

RHS = 4

Clearly ,

LHS ≠ RHS , thus the equation x + y = 4 doesn't passes through origin (0,0) .

★ Let's check whether x - y = 2 passes through origin (0,0) or not .

Putting x = 0 and y = 0 , we get ;

=> LHS = x - y

=> LHS = 0 - 0

=> LHS = 0

RHS = 2

Clearly ,

LHS ≠ RHS , thus the equation x - y = 2 doesn't passes through origin (0,0) .

★ Let's check whether x = 3y passes through origin (0,0) or not .

Putting x = 0 and y = 0 , we get ;

=> LHS = x

=> LHS = 0

=> RHS = 3y

=> RHS = 3×0

=> RHS = 0

Clearly ,

LHS = RHS , thus the equation x = 3y passes through origin (0,0) .

★ Let's check whether 3 = 2x + y passes through origin (0,0) or not .

Putting x = 0 and y = 0 , we get ;

LHS = 3

=> RHS = 2x + y

=> RHS = 2×0 + 0

=> RHS = 0 + 0

=> RHS = 0

Clearly ,

LHS ≠ RHS , thus the equation 3 = 2x + y doesn't passes through origin (0,0)

Hence ,

The linear equation x = 3y passes through origin.

Note :

If an equation doesn't contain any constant term , then it always passes through origin .

The linear equation x = 3y doesn't contain any constant term , thus it passes through origin .

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