Question

show that the straight lines x-2y+3=0 and 6x+3y+8=0 are perpendicular​

Answer 1

from eq 1

x=-3+2y

put the value of x in eq 2

6(-3+2y)+3y=-8

-18+12y+3y=-8

15y=10

y=2/3

Here

m1=1

m2=-3

m1×m2=-1

-3×1=-1

3

Hope it helps u❤

Answer 2

Step-by-step explanation:

slope of l1xl2=-1 then the line are perpendicular

l1slope =-2.

l2slope=1/2

l1×l2=-1

so line r perpendicular