Question

If a linear equation has solutions (-3,
3), (0,0) and (3, -3), then it is of the
form:
Select one:
A. -X +2 y = 0
B. y - x = 0
C. x + y = 0
D. -2x + y = 0​

Answer 1

Answer:

c) x+y =0

Step-by-step explanation:

it is the correct answer....okok

Answer 2

$Put \: (-3,3), \: (0,0) \: and \: (3,-3) \: in \:$

$Linear \: equations , we \:get$

$A. -x+2y = 0$

$put \: (-3,3) \: in \:the \: equation$

$- (-3) + 2\times 3 = 0$

$\implies 3 + 6 = 0$

$\implies 9 = 0 \: \red {( false) }$

$(-3,3) \:is \: not \:a \: solution$

$of \: given \: equation$

$B. y - x = 0$

$put \: (-3,3) \: in \:the \: equation$

$3 - (-3) = 0$

$\implies 3 + 3 = 0$

$\implies 6 = 0 \: \red {( false) }$

$(-3,3) \:is \: not \:a \: solution$

$of \: given \: equation$

$C. x+y = 0$

$i) put \: (-3,3) \: in \:the \: equation$

$-3 + 3 = 0$

$\implies 0 = 0 \: \green {( True) }$

$(-3,3) \:is \: not \:a \: solution$

$of \: given \: equation$

$ii) put \: (0,0) \: in \:the \: equation$

$0 + 0 = 0$

$\implies 0 = 0 \: \green {( True) }$

$(0,0) \:is \:a \: solution$

$of \: given \: equation$

$iii) put \: (3,-3) \: in \:the \: equation$

$3 - 3 = 0$

$\implies 0 = 0 \: \green {( True) }$

$(3,-3) \:is \: not \:a \: solution$

$of \: given \: equation$

$\therefore Option \green { (c)} \: is \: correct.$

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