Question

Write the equation of a line that is parallel to the line whose equation is 4x+3y=7 and also passes through the point (-6,2)?

Answer 1

The equation of a line that is parallel to the line whose equation is 4x+3y=7 and also passes through the point (-6,2) is: $y = -\frac{4}{3}x-6$

Step-by-step explanation:

Given equation of line is:

$4x+3y = 17$

We have to convert the given line in slope-intercept form to find the slope of the line

So,  Subtracting 4x from both sides

$4x+3y-4x = -4x+7\\3y = -4x+7$

Dividing both sides by 3

$\frac{3y}{3} = \frac{-4x+7}{3}\\y = \frac{-4x}{3} + \frac{7}{3}\\y = -\frac{4}{3}x+\frac{7}{3}$

The co-efficient of x is the slope of the line.

Let m1 be the slope of given line

$m_1 = -\frac{4}{3}$

As the line is parallel to given line, slope of both lines will be equal

Let m be the slope of line parallel to given line

m = $-\frac{4}{3}$

The slope-intercept form is:

$y=mx+b$

Putting the value of slope

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

Putting the point (-6,2) in the equation

$2 = -\frac{4}{3} (-6) +b\\2 = 4*2 + b\\2 = 8+b\\b = 2-8\\b = -6$

Putting the value of b

$y = -\frac{4}{3}x-6$

Hence,

The equation of a line that is parallel to the line whose equation is 4x+3y=7 and also passes through the point (-6,2) is: $y = -\frac{4}{3}x-6$

Keywords: Equation of line, slope

Learn more about equation of line at:

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