Question

A line through the origin intersects X 1, y =2 and x +y = 4, in A, B and C respectively.
such that OA.OB.OC = 8 V2. Find the equation of line.​

Answer 1

Answer:

y = x

Step-by-step explanation:

Let say line of Equation is

y = mx + c

Line passes through origin

so

0 = 0 + c

=> c = 0

Hence line equation

y = mx

intersect x = 1   at A

=> at A  y = m

A = ( 1 , m)

OA = √(1² + m²)

intersect y = 2   at B

=> at A  x = 2/m

B = (2/m , 2)

OB = √(4/m²) + 2² = 2√(m² + 1)  / m

x +y = 4

x + mx = 4

=> x = 4/(m + 1)

& y = 4m/(m+1)

C = (4/(m + 1) , 4m/(m+1))

OC = 4√(1/(m+1)²  + m²/(m+1)²)   = 4 √(m² + 1)  / (m + 1)

OA . OB . OC  = 8√2

=> √(1 + m²) * (2√(m² + 1)  / m ) * 4 √(m² + 1)  / (m + 1)  = 8√2

=> (1 + m² ) √(m² + 1)  = √2 m(m + 1)

Squaring both sides

=> (m² + 2m + 1)(m² + 1) = 2m²(m² + 2m + 1)

=> m⁴ + 2m³ + 2m² + 2m + 1  = 2m⁴ + 4m³ + 2m²

=> m⁴  + 2m³  - 2m - 1 = 0

=> (m - 1)(m³ + 3m² + 3m + 1) = 0

=> m = 1

y = x