Straight Lines

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

L 1 : 4 x + 3 y + 2 = 0  

L 2 : 3 x − 4 y − 1 1 = 0               

Since circle C touches the line L2 at Q intersection point L1 and L2 is (1, -2)

P lies of L1

∴ P ( x , − 1 3 ( 2 + 4 x ) )               

Now,

PQ = 5 ? (x – 1)2 + ( 4 x + 2 3 − 2 ) 2 = 2 5  

⇒ x = 4 , − 2             &n

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New answer posted•a year ago
S
a year ago

C : 4x2 + 4y2 – 12x + 8y + k = 0

? ( 1 , − 1 3 )               

Lies on or inside the C then

4 + 4 9 − 1 2 − 8 3 + k ≤ 0

⇒ k ≤ 9 2 9               

Now, circle lies in 4th quadrant centre

≡ ( 3 2 , − 1 )               

∴ r < 1 ⇒ 9 4 + 1 − k 4 < 1      

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

76. 

If point P be the junction between the lines

2x – 3y + 4 = 0 ______ (1)

3x + 4y – 5 = 0 ______ (2)

Solving (1) and (2) using 3 × (1) – 2 × (2) we get,

6x – 9y + 12 – (6x + 8y – 10) = 0

–17y + 22 = 0

y = 2217

And 2x = 3y– 4

=> 2x = 3 × 2217 – 4

x = 3317 – 2 = 33−3417 =&

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New answer posted•a year ago
S
a year ago

Kindly go through the solution

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New answer posted•a year ago
S
a year ago

Let A have the co-ordinate (x, o)

By laws of reflection

∠PAB = ∠ QAB = θ

And ∠ CAQ + θ = 90°

As normal is ⊥ to surface (x-axis)

⇒ ∠CAQ = 90° - θ

and ∠CAP = 90° + θ

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New answer posted•a year ago
S
a year ago

73. The given eqn of limes are.

9x + 6y – 7 = 0 ______ (1)

3x + 2y + b = 0 ______ (2)

Let P (x0, y0) be a point equidistant from (1) and (2) so

9x0 + 6y - 7 = ± 3 (3x0 + 2y0 + 6)

When, 9x0 + 6y0 – 7 = 3 (3x0 + 2y0 + 6)

⇒ 9x0 + 6y0 - 7 = 9x0 + 6y0 + 18

⇒ - 7 = 18 which in not true

So, 9x0 + 6y0 -

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New answer posted•a year ago
S
a year ago

72. The given eqn of the lines are.

x + y ? 5 = 0 _______ (1)

3x ? 2y + 7 = 0 ______ (2)

Given, sum of perpendicular distance of P (x, y)  from the two lines is always 10 .

The above eqn can be expressed as a linear combination Ax + By + C = 0 where A, B & C are constants representing a straight li

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New answer posted•a year ago