Find the equations of the lines, which cut-off intercepts on the axis whose sum and product are 1 and - 6 respectively

Solution:

Let the intercepts cut by the given lines on the axis be a and b.

It is given that

a + b = 1 ....(1)

ab = - 6 ....(2)

From (2), b = -6/a ... (3)

Substitute it in (1),

a + (-6/a) = 1

a² - a - 6 =0

(a - 3) (a + 2) = 0

a = 3 (or) a = -2

Substituting each of these in (3), we get, b = -2 (or) b = 3.

Thus,

a = 3 and b = - 2 or a = - 2 and b = 3

It is known that the equation of the line whose intercepts on the axis are a and b is

x/a + y/b = 1 or bx + ay - ab = 0

Case I: a = 3 and b = - 2

In this case, the equation of the line is - 2x + 3y + 6 = 0 ⇒ 2x - 3y = 6

Case II: a = - 2 and b = 3

In this case, the equation of the line is 3x - 2y + 6 = 0 ⇒ - 3x + 2y = 6

Thus, the required equations of the lines are 2x - 3y = 6 and - 3x + 2y = 6

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NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 3

Find the equations of the lines, which cut-off intercepts on the axis whose sum and product are 1 and - 6 respectively.

Summary:

The equations of the lines, which cut-off intercepts on the axis whose sum and product are 1 and - 6 respectively are 2x - 3y = 6 and - 3x + 2y = 6