Question

The area of triangle formed by the pair of straight lines (ax+by)2 -3(bx-ay)2=0 and ax+by+c =0
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Answer 1

Answer:

ax + by)? – 3(bx - ay)² = 0

= [ax + by + v3 (bx - ay)]

[ax + by - V3 (bx - ay)] = 0

= [(a+ 13 b)X + (-13 a + b)y]

[(a - 13 b)x + ( 13 a + b)y] = 0

: This equation represent the lines

(a + V3 b)x + (-13 a + b)y = 0 + (1),

(a - 13 b)x + ( 13 a + b)y = 0 + (2),

Let the given line be ax + by + c = 0 → (3)

If A is an angle between (1) and (3), then cosA

a(a + 3b)+b(-13a+b)

Va’ +b? Jla+ 136 )* +(-13a + b)*

a? + V3ab - 3ab + b?

Va? +b? Va? + 3b? + 213ab + 3a +b? - 213ab

a? + b2

Va? + b2 4a? +4b

BA = 60°

:: The angles between (1) and (3) are 60°, 120°

If B is an angle between (2) and (3), then cosB

a(a- 13b) +b(-13a +b)

Va? +b?/a- v3b)* +(53a+b)*

a? - V3ab + 3ab+b?

Va’ +62 Ja²3b2 - 213ab + 3a? + b2 + 2/3ab

a? + b2

1 1/2

Va? +b 4a? + 4b? 2

=

+

1

B = 60°

Step-by-step explanation:

ax + by)? – 3(bx - ay)² = 0

= [ax + by + v3 (bx - ay)]

[ax + by - V3 (bx - ay)] = 0

= [(a+ 13 b)X + (-13 a + b)y]

[(a - 13 b)x + ( 13 a + b)y] = 0

: This equation represent the lines

(a + V3 b)x + (-13 a + b)y = 0 + (1),

(a - 13 b)x + ( 13 a + b)y = 0 + (2),

Let the given line be ax + by + c = 0 → (3)

If A is an angle between (1) and (3), then cosA

a(a + 3b)+b(-13a+b)

Va’ +b? Jla+ 136 )* +(-13a + b)*

a? + V3ab - 3ab + b?

Va? +b? Va? + 3b? + 213ab + 3a +b? - 213ab

a? + b2

Va? + b2 4a? +4b

BA = 60°

:: The angles between (1) and (3) are 60°, 120°

If B is an angle between (2) and (3), then cosB

a(a- 13b) +b(-13a +b)

Va? +b?/a- v3b)* +(53a+b)*

a? - V3ab + 3ab+b?

Va’ +62 Ja²3b2 - 213ab + 3a? + b2 + 2/3ab

a? + b2

1 1/2

Va? +b 4a? + 4b? 2

=

+

1

B = 60°