Question

A straight line through the origin O meets the parallel lines 3x + 4y = 12 and 6x + 8y + 3 = 0 at
P and Q respectively. The ratio OP : OQ is
(B) 1:8
(C) 8:1
(D) 3:4
(A) 4:1​

Answer 1

Step-by-step explanation:

Given:Line4x+2y=9

Let (r

1

cosθ,r

1

sinθ) is on the line.

∴4r

1

cosθ+2r

1

sinθ=9

∴r

1

=

4cosθ+2sinθ

9

Let (−r

2

cosθ,−r

2

sinθ) lie on the line 2x+y+6=0

∴2r

2

cosθ+r

2

sinθ=6

∴r

2

=

2cosθ+sinθ

6

∴

r

2

r

1

= 2cosθ+sinθ6

=4cosθ+2sinθ9= 4cosθ+2sinθ

=9 × 62cosθ+ssin = 4:3

∴ OQ:OO = r:2r :1 = 4:3or 3:4

Answer 2

$\large{\underline{\underline{\sf{Given}}}}$

→ Two parallel straight line are given :-

$\sf{\implies \: 3x + 4y = 12 \: and \: 6x + 8y + 3 = 0}$

→ A line passing through origin intersects these two lines at P and Q point.

$\large{\underline{\underline{\sf{To \: Find}}}}$

→ The ratio of OP and OQ .

$\large{\underline{\underline{\sf{Solution}}}}$

Diagram refers to the attachment .

Let's assume that the ratio is k : 1.

So first of all we will solve the first equation by using the equation of line passing through the origin .

$\sf{\implies \: 3x+4y=12\: and\: y= mx}$

Putting the value of y in first equation .

$\sf{\implies 3x + 4(mx) = 12 }$

$\sf{\implies 3x + 4mx = 12 }$

$\sf{\implies x( 3 + 4m) = 12 }$

${\underline{\underline{\sf{\implies x = \frac{12}{3+4m} }}}}$

Now putting the value of x in equation first so that we can get the value of y also.

$\sf{\implies 3( \frac{12}{3 + 4m} ) + 4y = 12 }$

Taking 4 common from both sides

$\sf{\implies 4 [ ( \frac{9}{3 + 4m} + y ] = 4[ 3 ] }$

$\sf{\implies \frac{9 + 3y + 4ym }{ 3+4m} = 3 }$

$\sf{\implies 9 + 3y + 4ym = 9 + 12m }$

$\sf{\implies y(3 + 4m) = 9-9+12m }$

${\underline{\underline{\sf{\implies y = \frac{12m}{3+4m} }}}}$

${\boxed{\sf{Coordinates\: of \: P( \frac{12}{3+4m} , \frac{12m}{3+4m}}}}$

Now we will solve equation second by the equation of line passing through origin .

$\sf{\implies 6x + 8y = -3 \: and \: y = mx }$

Putting the value of y in equation second.

$\sf{\implies 6x + 8(mx) = -3 }$

$\sf{\implies 6x + 8mx = -3 }$

$\sf{\implies x(6+8m) = -3 }$

${\underline{\underline{\sf{\implies x = \frac{-3}{6 + 8m} }}}}$

Now putting the value of x in equation second so that we can get the value of y also.

$\sf{\implies 6[ \frac{-3}{6 + 8m} ] + 8y = -3 }$

$\sf{\implies [ \frac{-18}{6+8m} ] + 8y = -3 }$

$\sf{\implies \frac{- 18 + 48y + 64my}{6+8m} = -3 }$

$\sf{\implies -18 + 48y + 64my = -18 - 24m }$

$\sf{\implies y( 48 + 64m) = -24m }$

${\underline{\underline{\sf{\implies y = \frac{-24}{ 48 + 64 m } }}} }$

Taking 8 common from RHS fraction .

${\underline{\underline{\sf{\implies y = \frac{-3m}{6 + 8m} }}}}$

$\sf{Coordinates\: of \: Q ( \frac{-3}{6+8m} , \frac{-3m}{6+8m} }$

Now we know co-ordinate of origin are O( 0,0)

Using intersection formula.

$\sf{\implies Ox = \frac{m_1 ( x_2) + m_2(x_1)}{k+1} }$

$\sf{\implies Ox = \frac{k ( \frac{-3}{6+8m} + 1( \frac{12}{3+4m}}{k+1} }$

→ 0 = [-3k/6+8m + 12/3+4m]/[k+1]

→ 0 = [-3k/2 + 12]

→ -12 = -3k/2

→ -24 = -3k

→ 24/3 = k

→ 8/ 1 = k/1

So Ratio is 8 : 1