Question

$please \: answer \: with \: explanation \: .best \: answer \: = 30 \: thanks \\ \\ + brainliest \: . \: dont \: answer \: me \: \: if \: u \: dont \: know \: the \: answer$
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Answer 1

Step-by-step explanation:

hope it will help you mark as brainlyst .

Answer 2

$\huge\underline\bold\red{AnswEr}$

☞︎︎︎A line AB forms a triangle with coordinate axes of area 54√3 [ I think you did mistake in typing area of triangle may be 54√3 , for better calculation , btw you can do with your mentioned too . But I am using 54√3 unit²] square unit and also perpendicular drawn from the origin to the line makes an angle of 60° with x -axis as shown in figure.

area of triangle = 1/2 × height × base

see attachment, height = a and base = b

then, area of triangle = 1/2ab = 54√3 ⇒ab = 108√3 ------(1)

Now, Let P is the length of perpendicular drawn from origin to line

Then, cos60° = P/b ⇒b = P/cos60° = 2P

cos30° = P/a ⇒a = P/cos30° = 2P/√3

Now, put a and b value in equation (1),

4P²/√3 = 108√3 ⇒P² = 108 × 3

P = ±18 , but we have to take only P = 18 because triangle is in 1st quadrant .

a = 2P/√3 = 36/√3 = 12√3

b = 2P = 36

so, equation of line : cos60°x + sin60° y= P

x + √3y - 36 = 0

Hence, equation of line : x + √3y -36 = 0