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Question

 

Find the equation of a line through (2,2) and forming with the axes a triangle of area 9.?

Answer 1

Let 'a' and 'b' be the intercept of the line on x and y axis.

Then the equation of the line is x/a+y/b = 1----------->(1)

Since this passes through the point P(2, 2)

2/a+2/b = 1

(2b + 2a) / (ab)  = 1          a, b LCM is ab

Multiply each side by (ab).

[(2b + 2a) / (ab) ](ab) = 1 (ab)

Simplify

2a + 2a = ab

Take out common factor 2.

2(a + b) = ab

Divide each side by 2

[2(a + b)] / 2 = (ab) / 2

a + b = ab/2 --------------> (2)

But the area of OAB, where O is (0, 0), A (a,0) and B (0,b) is  ab/2 sqre.unit

Given that area is 9 sqre.unit

So,  (ab)/2 = 9

Multiply each side by 2.

[(ab)/2] 2 = 9(2)

ab = 18

Assume a>0 , b>0. Then ab =18 ----------------> (3).

Substitute ab = 18 in the equation(2).

(2)⇒(a + b) = 18/2

a + b = 9 ----------------> (4).

Recall: [ (a - b)² = (a + b)² - 4ab ]

From equation(3) is ab = 18 and equation(4) is a + b = 9

Substitute a + b = 9 and ab = 18 above the formula.

Then (a - b)² = (9)² - 4(18)

Simplify

(a - b)² = 81-72

(a - b)² = 9

Take square root each side.

√[(a - b)²] = √3²

Recall: √A² = A

a - b = 3 ------------->(5)

From (4) and (5) solving for,

                              a + b = 9

                              a - b = 3

        (add)___________________

                        2a = 12

Divide each side by 2.

2a / 2 = 12 / 2

a = 6

Substitute a = 6 in the equation(4).

From equation (4) is a + b = 9

Then 6 + b = 9

Subtract 6 from each side.

6 + b - 6 = 9 - 6

b = 3

Substitute a = 6 and b = 3in the equation(1).

From equation (1) is x/a+y/b = 1

x / 6 + y / 3 = 1

(x + 2y) / 6 = 1                Here 3, 6 LCM is 6

Multiply each side by 6.

[(x + 2y) / 6] 6 = 1(6)

x + 2y = 6

Subtract 6 from each side.

x + 2y - 6 = 6 - 6

x + 2y - 6 = 0

There fore the equation is x + 2y - 6 = 0

Answer 2

Equation of the line is y = mx + b.

Line passes through (2, 2).

2 = 2m + b ->(1)

Line intersects x-axis at y = 0.

0 = mx + b => mx = -b => x = -b/m

Line intersects y-axis at x = 0.

y = m(0) + b => y = b

Area of the triangle is 1/2 * b * h = 9

1/2 * -b/m * b = 9

-b² / 2m = 9

18m = -b²

m = -b² / 18   -> (2)

Substitute m value in (1).

2 = 2m + b

2 = 2(-b² / 18) + b

2 = -b²/9 + b

18 = -b² + 9b

b² - 9b + 18 = 0

b² - 6b - 3b + 18 = 0

b(b - 6) - 3(b - 6) = 0

(b - 6)(b - 3) = 0

b - 6 = 0 or b - 3 = 0

b = 6 or b = 3

Substitute b values in equation (2).

If b = 6, then m = -b² / 18 = -36/18 = -2.

If b = 3, then m = -b² / 18 = -9/18 = -1/2.

Equations of the line are y = -2x + 6 or y = -1/2 x + 3.

2x + y = 6 or x +2y = 6.