Question

MPLE 12 Fond the equation of the line midway between the parallel lines
9x + y - 7=0 and 3x +2y+6=0.​

Answer 1

Answer:

$\huge\underline{ \underline{ \red{your \: \: answer}}}$

Step-by-step explanation:

$\sf\pink{The \: given \: lines \: are} \\ \sf\pink{9x + 6y - 7 = 0} \\ \sf\pink{ i.e., 3x + 2y - 7/3 = 0 ....(i) } \\ \sf\pink{ and \: 3x + 2y + 6 = 0 ....(ii) } \\ \sf\blue{ Let, \: the \: parallel \: line \: is \: - } \\ \sf\red{ 3x + 2y + k = 0 .....(iii) } \\ \sf\red{ \:</p><p>Since \: (iii) \: no \: line \: passes \: midway \: between } \\ \sf\red{ \:(i) and (ii) the distance between } \\ \sf\red{ (iii) and (i) (iii) and (ii) are \: equal} \\ \sf\orange{ thus = > \frac{ |(k + \frac{7}{3} )| }{ \sqrt{13} } = \frac{ |(k - 6)| }{ \sqrt{13} } } \\ \sf\orange{ = > {(k + \frac{7}{3}) }^{2} = {(k - 6)}^{2} } \\ \sf\orange{ = > \frac{14k}{3} + \frac{49}{9} = - 12k + 36 } \\ \sf\orange{ = > \frac{50k}{3} = \frac{279}{9} } \\ \sf\orange{ = > k = \frac{11}{6} } \\ \sf\purple{ \:Therefore, the \: required \: parallel \: line \: is} \\ \sf\purple{ \:3x + 2y + 11/6 = 0} \\ \sf\purple{ \:i.e., 18x + 12y + 11 = 0} \\ \large\boxed{ \fcolorbox{pink}{red}{hope \: it \: help \: you}}$

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

The given parallel lines are:

→9x + 6y - 7 = 0 ......(1)

→3x + 2y + 6 = 0 .......(2)

$\sf{Multiplying \: equation \: (2) \: by \: 3}$

9x + 6y + 18=0 ......(3)

Let 9x + 6y + a = 0 .....(4) be the line parallel.

Which are equidistant from the line (1) and(2).

The distance between (1)and(4) is equal to The distance between (3)and(4)

a + 7 = a - 18

→a + 7 = -(a - 18)

→2a = 11

→a = 11/2 .......(5)

what we need is only a value .

Putting (4) in (5),

9x + 6y + 11/2 = 0

18x + 12y + 11 = 0 is the required line.