Question

Find a relation between x and y such that the point (x, y) is equidistant from points (7, 1) and (3, 5).

Answer 1

$\bold{\huge{\fbox{\color{Maroon}{♡Hello♡}}}}$

Answer :

(3, 7), (6, 5) and (15, –1).✔️

=>Let the points be A (15, –1), B (6, 5) and C (3, 7)

Distance of AB

⇒ AB = √ (6 – 15)2 + (5 – (–1))2

⇒ AB = √ (–9)2 + (6)2

⇒ AB = √ (81 + 36)

⇒ AB = √ 117 = √ 3 × 3 × 13

⇒ AB = 3√13

Distance of BC

⇒ BC = √ (3 – 6)2 + (7 – 5)2

⇒ BC= √ (3)2 + (2)2

⇒ BC = √ (9 + 4)

⇒ BC= √ 13

Distance of AC

⇒ AC = √ (3 – 15)2 + (7 – (–1))2

⇒ AC = √ (3 – 15)2 + (7 + 1)2

⇒ AC= √ (–12)2 + (8)2

⇒ AC = √ (144 + 64)

⇒ AC= √ 208 = √ 4 × 4 × 13

⇒ AC = 4√13

i.e. AB + BC = AC

⇒ 3√13 + √13 = 4√13

$\bold{\small{\fbox{\color{Black}{∴ A, B and C are collinear}}}}$

$\bold{\huge{\fbox{\color{Pink}{♡Thanks♡}}}}$

Answer 2

$\bold{Answer..}$

(3, 7), (6, 5) and (15, –1)

=>Let the points be A (15, –1), B (6, 5) and C (3, 7)

Distance of AB

⇒ AB = √ (6 – 15)2 + (5 – (–1))2

⇒ AB = √ (–9)2 + (6)2

⇒ AB = √ (81 + 36)

⇒ AB = √ 117 = √ 3 × 3 × 13

⇒ AB = 3√13

Distance of BC

⇒ BC = √ (3 – 6)2 + (7 – 5)2

⇒ BC= √ (3)2 + (2)2

⇒ BC = √ (9 + 4)

⇒ BC= √ 13

Distance of AC

⇒ AC = √ (3 – 15)2 + (7 – (–1))2

⇒ AC = √ (3 – 15)2 + (7 + 1)2

⇒ AC= √ (–12)2 + (8)2

⇒ AC = √ (144 + 64)

⇒ AC= √ 208 = √ 4 × 4 × 13

⇒ AC = 4√13

i.e. AB + BC = AC

⇒ 3√13 + √13 = 4√13