Question

Find the equation of line
perpendicular to line 3x-2y +5=0
and passing through the point
1,3)​

Answer 1

Answer:

2x+3y-11=0

Step-by-step explanation:

Since the line is perpendicular to 3x-2y +5=0, product of both of its slopes will be -1.

Slope of 3x-2y +5=0:

2y=3x+5

y=(3/2)x + (5/2), Comparing this with y=mx+c, slope is 3/2.

Now let the slope of perpendicular line be m then,

m*(3/2)=-1

m=(-2/3)

Hence the equation of perpendicular line would be,

y=(-2/3)x+c, Also the line is passing through (1,3) hence,

3=(-2/3)*1 +c

c=3+(2/3)= 11/3

Hence the equation of line would be,

y=(-2/3)x + (11/3)

3y=-2x+11

2x+3y-11=0

Cheers!

Answer 2

Answer:

Answer:-2

Step-by-step explanation:

Let x=1 & y=3

Put this value in this equation

3x-2y+5=0

As we know x=1 & y=3

3(1)-2(3)+5=0

3-6+5=0

8-6=0

2=0