Question

If a point P moves such that the distance from the point A(1,1) and the line x+y+2=0 are equal then the locus of P is equal to

Answer 1

Let the point P be (h, k), Now it is given that the point P is always equidistant from the point A(1, 1) and the line x + y + 2 = 0

This can be expressed as,

⇒ AP = Distance of A from line x + y + 2 = 0

The distance between the points A and P can be found using the distance formula, so let's try to find the distance of the point P from the line.

We know, The distance of a point (x₁ , y₁) from a line ax + by + c = 0 is given by,

⇒ D = | { ax₁ + by₁ + c } / { √(a² + b²) } |

Substituting the values in the given formula, we get

⇒ D = | { 1×h + 1×k + 2 } / { √(1 + 1) } |

⇒ D = | (h + k + 2) / √2 |

Square both sides to remove the modulus, because after squaring the value will always be positive, so no need of modulus

⇒ D² = (h + k + 2)² / 2 ...(i)

Now,

⇒ AP² = D²

⇒ (1 - h)² + (1 - k)² = (h + k + 2)² / 2

⇒ 2(1 + h² - 2h + 1 + k² - 2k) = h² + k² + 4 + 2hk + 4k + 4h

⇒ 2 + 2h² - 4h + 2 + 2k² - 4k = h² + k² + 4 + 2hk + 4k + 4h

⇒ 2h² - h² + 2k² - k² - 4h - 4h - 4k - 4k - 2hk = 4 - 2 - 2

⇒ h² + k² - 8h - 8k - 2hk = 0

Replacing h with x and k with y, we get

⇒ x² + y² - 8x - 8y - 2xy = 0

Answer 2

Let the point P be (h, k)

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━

⠀

→ D = | { ax₁ + by₁ + c } / { √(a² + b²) } |

→ D = | { 1 × h + 1 × k + 2 } / { √(1 + 1) } |

→ D = | (h + k + 2)/√2 |

→ D² = (h + k + 2)²/2 ...(i)

Now,

→ AP² = D²

→ (1 - h)² + (1 - k)² = (h + k + 2)²/2

→ 2(1 + h² - 2h + 1 + k² - 2k) = h² + k² + 4 + 2hk + 4k + 4h

→ 2 + 2h² - 4h + 2 + 2k² - 4k = h² + k² + 4 + 2hk + 4k + 4h

→ 2h² - h² + 2k² - k² - 4h - 4h - 4k - 4k - 2hk = 4 - 2 - 2

→ h² + k² - 8h - 8k - 2hk = 0

→ x² + y² - 8x - 8y - 2xy = 0