Question

Value of lamda so that point (lamda,lamda²) lies between the lines x+2y=3, x+2y=-3 is
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Answer 1

Given that the point (λ, λ²) lies between the lines x + 2y - 3 = 0 and x + 2y + 3 = 0.

So the distance of the point (λ, λ²) from the straight line x + 2y - 3 = 0 is

= |λ + 2λ² - 3| / {√(1² + 2²)} units

= |λ + 2λ² - 3| / (√5) units

and the distance of the point (λ, λ²) from the straight line x + 2y + 3 = 0 is

= |λ + 2λ² + 3| / {√(1² + 2²)} units

= |λ + 2λ² + 3)| / (√5) units

∵ the point (λ, λ²) lies between the given lines,

|λ + 2λ² - 3| / (√5) = |λ + 2λ² + 3| / (√5)

or, λ + 2λ² - 3 = ± (λ + 2λ² + 3)

When we take the positive + sign, we cannot determine the value of λ.

We take the negative - sign:

λ + 2λ² - 3 = - (λ + 2λ² + 3)

or, λ + 2λ² - 3 = - λ - 2λ² - 3

or, λ + 2λ² = 0

or, λ (2λ + 1) = 0

Either λ = 0 or, 2λ + 1 = 0

i.e., λ = 0, - 1/2

Therefore λ = 0, - 1/2

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Answer 2

The value of the point λ is either 0 or - 1/2

Step-by-step explanation:

The distance between line and point is given by the formula:

d = |ax₁ + by₁ + c|/√(a² + b²)

The distance between the point (λ, λ²) and line x + 2y - 3 = 0 is

a = 1; b = 2; c = -3; x₁ = λ; y₁ = λ²

On subsituting the values, we get,

d = |λ + 2λ² - 3|/√(1² + 2²)

∴ d = |λ + 2λ² - 3|/(√5) units

The distance between the point (λ, λ²) and line x + 2y + 3 = 0 is

a = 1; b = 2; c = 3; x₁ = λ; y₁ = λ²

d = |λ + 2λ² + 3|/√(1² + 2²)

∴ d = |λ + 2λ² + 3)|/(√5) units

Now, since, the point lies between the lines,

|λ + 2λ² - 3|/(√5) = |λ + 2λ² + 3)|/(√5)

∴ λ + 2λ² - 3 = ± (λ + 2λ² + 3)

Since, there is no possibility with positive sign,

λ + 2λ² - 3 = - (λ + 2λ² + 3)

λ + 2λ² - 3 = - λ - 2λ² - 3

λ + 2λ² = 0

λ (2λ + 1) = 0

λ = 0 and 2λ + 1 = 0

∴ λ = 0, - 1/2