Question

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​

Answer 1

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 ≡

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

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

$\implies 4y+20=5x+20$

$\implies 4y = 5x$