Question

Find the equation of lines which makes intercepts on the axis whose sum is 1 and product is -6

Answer 1

Answer:

2x - 3y = 6

and

-3x + 2y = 6

Step-by-step explanation:

First, we need two numbers whose sum is 1 and product is -6.

These are the roots of the quadratic equation

x² - (sum of roots) x + (product of roots) = 0

=> x² - x - 6 = 0

=> ( x - 3 ) ( x + 2 ) = 0

=> x = 3 or x = -2.

So the intercepts are at 3 and at -2.

Line 1 :  x-intercept at 3 and y-intercept at -2

x/3  +  y/(-2)  = 1     <=>  2x - 3y = 6

Line 2 :  x-intercept at -2 and y-intercept at 3

x/(-2)  +  y/3 = 1    <=>  -3x + 2y = 6

Answer 2

• When (a = 3), (b = -2) = 2x - 3y - 6 = 0
• When (a = -2), (b = 3) = 3x - 2y + 6 = 0

Step-by-step explanation:

Let,

• The intercepts of a line are a, b respectively.

Given,

• a + b = 1 -------- (1)
• ab = -6 -------- (2)

By assuming a and b as roots of the equation,

⟼ x² - x (sum of the roots) + product of the root = 0

⟼ x² - x (1) - 6 = 0

⟼ x² - x - 6 = 0

By factorizing,

⟼ (x - 3) (x + 2) = 0

⟼ x = 3 or x = -2

Hence, the intercepts are 3 or -2

Case ❶ :

When a = 3, b = -2,

⟼ (x / a) + (y / b) = 1

⟼ (x / 3) + (y / -2) = 1

⟼ x / 3 - y / 2 = 1

⟼ 2x - 3y / 6 = 1

⟼ 2x - 3y = 6

⟼ 2x - 3y - 6 = 0

Case ❷ :

When a = -2, b = 3,

⟼ (x / a) + (y / b) = 1

⟼ (x / -2) + (y / 3) = 1

⟼ x / -2 + y / 3 = 1

⟼ -3x + 2y / 6 = 1

⟼ -3x + 2y = 6

⟼ -3x + 2y - 6 = 0

⟼ 3x - 2y + 6 = 0

• Hence, When (a = 3), (b = -2) = 2x - 3y - 6 = 0
• When (a = -2), (b = 3) = 3x - 2y + 6 = 0