Question

If the point (a.a) falls between the lines |x+y|=2 then

Answer 1

Distance of a straight line from a given point:

    Let, (p, q) be a point given and a straight line is

    Ax + By + C = 0 ..... (1)

Then the distance of the line (1) from the point is given by

d = $\frac{|Ap+Bq+C|}{\sqrt{A^{2}+B^{2}}}$ units

The complete question is:

    If the point (a, a) lies between the lines | x + y | = 2, find the value of a.

Solution:

The given lines are

    | x + y | = 2

So the two lines are

x + y = 2 & x + y = - 2

The given point is (a, a)

Then the distance of the line x + y = 2 from the point (a, a) is

d₁ = $\frac{|a+a-2|}{\sqrt{1^{2}+1^{2}}}$ units

= $\frac{|2a-2|}{\sqrt{2}}$ units

and the distance of the line x + y = - 2 from the point (a, a) is

d₂ = $\frac{|2a+2|}{\sqrt{1^{2}+1^{2}}}$ units

= $\frac{|2a+2|}{\sqrt{2}}$ units

ATQ, d₁ = d₂

  or, d₁² = d₂²

  or, | 2a - 2 |² = | 2a + 2 |²

  or, (2a - 2)² = (2a + 2)²

  or, 4a² - 8a + 4 = 4a² + 8a + 4

  or, 16a = 0

  or, a = 0

∴ the value of ‘a’ is 0 and thus the point is the origin (0, 0).