Question

Draw the following of x+y=6 which intersects the x-axis and the Y- axis at A and B respectively. find the length of seg AB. also find the area of ∆AOB, where O is the point of origin​

Answer 1

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GiveN :

$\rightsquigarrow \sf linear \: \: {eq}^{n} \: \: is \: \: \: x + y = 6$

$\rightsquigarrow \sf \: it \: \: intersects \: \: x -axis \: \: at \: \: pt = A( \: x \: , \: 0 \: )$

$\rightsquigarrow \sf \: it \: \: intersects \: \: y -axis \: \: at \: \: pt = B( \: 0 \: , \: y \: )$

To FinD :

$\rightsquigarrow \sf length \: \: of \: \: {AB}$

$\rightsquigarrow \sf area \: \: of \: \: \triangle \: {AOB}$

SolutioN :

x+y=6 intersects x-axis :

$\sf y = 0 \: \: when \: \: it \: \: intersects \: \: x \: axis$

$\sf Hence, \: \: putting \: \: y=0$

$\dashrightarrow \sf \: x \: + 0 = 6$

$\dashrightarrow \sf { {{ x \: = 6}}}$

$\dashrightarrow \sf { \underline{\underline{ pt \: = (6 \: , \: 0)}}}$

x+y=6 intersects y-axis :

$\sf x = 0 \: \: when \: \: it \: \: intersects \: \: y \: axis$

$\sf Hence, \: \: putting \: \: x=0$

$\dashrightarrow \sf \: 0 \: + y = 6$

$\dashrightarrow \sf { {{ y \: = 6}}}$

$\dashrightarrow \sf { \underline{\underline{ pt \: = (0 \: , \: 6)}}}$

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~DISTANCE

$\dashrightarrow \sf \: AB = \sqrt{ {(0 - 6)}^{2} + {(6 - 0)}^{2} }$

$\: \: \therefore \: { \underline{ \boxed{\sf \: AB = 6\sqrt{2}}}}$

________________

~AREA

distance of pt A from pole is 6 unit

and distance of pt B from pole is 6 unit

$\dashrightarrow \sf \triangle \: AOB = \frac{1}{2} \times 6 \times 6 \: \:$

$\: \: \: \therefore \underline{ \boxed{ \sf \triangle \: AOB = 18}}$

RequireD ConcepT :

$\circ \sf \: \: Area \: \: of \: \: a \: \: triangle \: = \frac{1}{2} \times base × hight$

$\circ \sf \: \: Distance \: \: of \: \: \: two \: \: pts \: = \sqrt{(y_2-y_1)²+(x_2-x_1)²}$

__________________

HOPE THIS IS HELPFUL...

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Answer 2

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