Question

Find the equation of the line parallel to x+3y = 4 and passing through the point (2, 5)

Answer 1

Answer:

$\displaystyle y-5=-\frac{1}{3}(x-2)$

Step-by-step explanation:

Equation of a Line

Two lines are parallel if they have the same slope. We are given the equation of the line:

x+3y=4

To find the slope of this line, we must solve for y:

$\displaystyle y=-\frac{1}{3}x+\frac{4}{3}$

The slope of this line is:

$\displaystyle -\frac{1}{3}$

The equation of the new line has the same slope. To find its equation, we use the point-slope form of the line:

y-k=m(x-h)

Where (h,k)=(2,5). Thus:

$\boxed{\displaystyle y-5=-\frac{1}{3}(x-2)}$

Answer 2

Answer:

The equation of the parallel line is x + 3y = 17

Step-by-step explanation:

Parallel lines have the same slopes and different y-intercept

The form of the equation y = m x + b, where

• m is the slope of the line

• b is the y-intercept

∵ The equation of the given line is x + 3y = 4

→ We must put it in the form above to find its slope

→ Subtract x from both sides to move x to the right side

∵ x - x + 3y = 4 - x

∴ 3y = 4 - x

→ Divide both sides by 3 to make the coefficient of y = 1

∴ $\frac{3y}{3}=\frac{4}{3}-\frac{1}{3}x$

∴ y = $\frac{4}{3}$ -  $\frac{1}{3}$ x

→ Compare it with the form of the equation above

∴ m = $-\frac{1}{3}$

∵ Parallel lines have the same slopes

∴ The slope of the parallel line is $-\frac{1}{3}$

→ Put it in the form of the equation

∴ y =  $-\frac{1}{3}$ x + b

→ To find b substitute x and y in the equation by the coordinates

   of a point on the line

∵ The line passes through the point (2, 5)

∴ x = 2 and y = 5

→ Substitute them in the equation

∴ 5 =  $-\frac{1}{3}$ (2) + b

∴ 5 =  $-\frac{2}{3}$ + b

→ Add  $\frac{2}{3}$  to both sides

∴ $\frac{17}{3}$ = b

→ Substitute it in the form of the equation above

∴ y = $-\frac{1}{3}$ x + $\frac{17}{3}$

→ Multiply each term by 3

∴ 3y = - x + 17

→ Add x to both sides

∴ x + 3y = 17

∴ The equation of the parallel line is x + 3y = 17