Question

Find an equation of the line that is parallel to the graph of 2x+3y=5 and contains the point (3, −1).

Answer 1

Answer:

A) Equation of line is 3X - 2Y = 11

B) Equation of line is Y = X + 1  

C) Equation of line is Y - 7 = 0

Equation of line , point (3 , -1) and slop $\frac{3}{2}$ is 3X - 2Y = 11

Step-by-step explanation:

Given equation of line as  2x + 3y = 5

slove of this line (m1) = - $\frac{2}{3}$

Now another line passes through point  (3 , -1) having slop (m2)

Both lines are parallel to each other, i.e (m1) × (m2) = - 1

So slop of second line (m2 ) = $\frac{-1}{m1}$

                                      (m2) = $\frac{-1}{(\frac{-2}{3} } )$

                                      (m2) = $\frac{3}{2}$

Now eq of line with point (3 , -1) and slop (m2)

Y- y1 = (m2)( X - x1)

Or, Y +1 = $\frac{3}{2}$ (X - 3)

i.e 2Y + 2= 3X - 9

∴ Equation of line is 3X - 2Y = 11     Answer

B) Given points as ( 5,6) and (3 , 4)

Slop (m) = $\frac{y2 - y1}{x2 - x1}$ = $\frac{4 - 6}{3 - 5}$ = 1

So the equation of line is

Y- y1 = (m)( X - x1)

Y - 6 = (1) (X - 5)

i.e Equation of line is Y = X + 1   Answer

C) Given points as ( 0, 7)  , (0 , -8)

Slop (m) with co-ordianate =  $\frac{y2 - y1}{x2 - x1}$

Or,                                        =  $\frac{-8 - 7}{0 - 0}$

                                            =0

Hence equation of line is Y - 7 = 0     Answer