A ray of light coming from the point (2,2√3) is incident at an angle 30° on the line x=1 at the point A. The ray gets reflected on the line x=1 and meets x-axis at the point B. Then, the line AB passes through the point:
A ray of light coming from the point (2,2√3) is incident at an angle 30° on the line x=1 at the point A. The ray gets reflected on the line x=1 and meets x-axis at the point B. Then, the line AB passes through the point:
m (AP)=tan60°=√3. y-2√3=√3 (x-2). At x=1, y=√3. A= (1, √3).
m (AB)=tan120°=-√3. y-√3=-√3 (x-1) ⇒ √3x+y=2√3. (3, -√3) satisfies
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...(1)
–2α + β = 0 …(2)
Solving (1) and (2)
a =
|z| = 0 (not acceptable)
|z| = 1
|z|2 = 1
Given : x2 – 70x + l = 0
->Let roots be a and b
->b = 70 – a
->= a (70 – a)
l is not divisible by 2 and 3
->a = 5, b = 65
->
z1 + z2 = 5
⇒ 20 + 15i = 125 – 15z1z2
⇒ 3z1z2 = 25 – 4 – 3i
3z1z2 = 21– 3i
z1⋅z2 = 7 – i
(z1 + z2)2 = 25
= 11 + 2i
&nb
a = 1 > 0 and D < 0
4 (3k – 1)2 – 4 (8k2 – 7) < 0
K = 3
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Maths NCERT Exemplar Solutions Class 11th Chapter Four 2025
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