Math, asked by yash81830, 4 months ago

find the equation of line passing through the point of intersection of 2x+y+6=0 and 3x+5y-15=0 and parallel to the line 5x+6y+3=0​

Answers

Answered by gopalpvr
0

Step-by-step explanation:

2x+y= -6-------1

3x+5y=15-------2

equation1 × 5-equation2

10x+5y=-30

3x+5y= -15

__________

7x = -15

X= -15/7

from 1

2(-15/7)+y=-6

Y= -6+30/7

=(-42+30)/7

= -12/7

point of intersection (-15/7, -12/7)

slope of 5x+6y+3=0 is m= _5/6

equation of required line =( y-y1) = m(x-x1)

y-(-12/7) =-5/6 (x-(-15/7))

(7y+12)/7= -5/6(7x+15)/7

6(7y+12)=-5(7x+15)

42y+72= -35x - 75

35x+42y+147= 0

equation of the required line = 35x+42y+147= 0

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