Test: Distance Between Lines - JEE MCQ

• The given equation is 
• x +√3y = 2 
• On dividing both sides by  
• [(−1)² + (√3)²]^1/2 = √4 = 2
•  = x/2 + √3y/2 = 2/2 
• Therefore , the equation becomes
• = (1/2)x + (√3/2)y = 1
• = x cos 60° + y sin 60° = 1
• Normal form of any line is : x cosw + y sinw = p
• w = 60°

The foot of perpendicular can be found by equating the distance between the two points and the distance between point and line. This is found to be (68/25,-49/25).

2(x-3) = 3(y+5)
2x - 6 = 3y + 15
2x - 3y = 21
2x - 3y - 21 = 0
Using distance formula = |ax₁ + by₁ + c|/(a² + b2)^1/2
= (6-9-21)/((2)² + (3)²)^1/2
= 24/(13)^1/2

In Δ PAM & Δ PAN, angle (PAM) = angle (PAN) (Since AP bisects the angle (BAC)) angle (AMP) = angle (ANP) = 900
PA = PA (common)
By AAS congruence, Δ PAM and Δ PAN are congruent. 
MP = NP = 2 cm

√3x + y - 8
√3x + y = 8
Dividing by √[(√3)2 + (1)2]
= √[3+1]
= √4
= 2
√3x/2 + y/2 = 8/2
√3x/2 + y/2 = 4
x(√3/2) + y(1/2) = 4.....(1)
Normal form of any line : xcos w + ysin w = p....(2)
Comparing (1) and (2)
p = 4

4x - 3y + 5 = 0,  4x - 3y + 15 = 0
A = 4,  B = -3   c₁ = 5,   c₂ = 15
|c₁ - c₂|/[A² + B²]^1/2
= |5 - 15|/[(4)² + (-3)²]^½
= 10/5
= 2