Math, asked by anuriu1996, 2 months ago

8. Find the eqation of the straight line passing through the point of interscetion of the line x+y+9=0 and 3x-2y+2=0 and is perpendicular to the line 4x+5y+1=0​

Answers

Answered by senboni123456
0

Answer:

Step-by-step explanation:

Given lines are x+y+9=0 and 3x-2y+2=0

To find the point of contact, multiply the first equation by 3 and subtract from the second equation i.e.,

(3x+3y+27)-(3x-2y+2)=0

⇒ 5y+25=0

⇒ y= -5

Putting y = -5 in the first equation, we get,

x = -4

So, the required line is passing through  (-4,-5)

It is also given that the required line is perpendicular to the line 4x+5y+1=0

⇒5y = -4x-1

 \implies y=-\frac{4}{5}-\frac{1}{5}\\

So, slope of required line ,  m =\frac{5}{4}

Now, equation of line ≡

 (y+5)=m(x+4)

\implies (y+5)=\frac{5}{4}(x+4)

\implies 4y+20=5x+20

\implies 4y = 5x

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