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Find the equation of the lines parallel and perpendicular to the line 5x+2y=12 through the point (-2,3)

Respuesta :

Answer:

The equation of line parallel to given line and passing through points        ( - 2 , 3 ) is 5 x + 2 y + 4 = 0

The equation of line perpendicular to given line and passing through points ( - 2 , 3 ) is 2 x - 5 y + 19 = 0

Step-by-step explanation:

Given equation of line as :

5 x + 2 y = 12

or, 2 y = - 5 x + 12

or , y = [tex]\frac{-5}{2}[/tex] x + [tex]\frac{12}{2}[/tex]

Or, y = [tex]\frac{-5}{2}[/tex] x + 6

∵ Standard equation of line is give as

y = m x + c

Where m is the slope of line and c is the y-intercept

Now, comparing given line equation with standard eq

So, The slope of the given line = m = [tex]\frac{-5}{2}[/tex]

Again,

The other line if passing through the points (- 2 , 3 ) And  is parallel to given line

So, for parallel lines condition , the slope of both lines are equal

Let The slope of other line = M

So,  M = m = [tex]\frac{-5}{2}[/tex]

∴ The equation of line with slope M and passing through points ( -2 , 3) is

y = M x + c

Now , satisfying the points

So, 3 = [tex]\frac{-5}{2}[/tex] × ( - 2 ) + c

or, 3 =  [tex]\frac{10}{2}[/tex] + c

Or, 3 = 5 + c

∴  c = 3 - 5 = - 2

c = - 2

So, The equation of line with slope  [tex]\frac{-5}{2}[/tex]  and passing through points ( -2 , 3)

y =  [tex]\frac{-5}{2}[/tex] x - 2

or, 2 y = - 5 x - 4

I.e 5 x + 2 y + 4 = 0

Similarly

The other line if passing through the points (- 2 , 3 ) And  is perpendicular  to given line

So, for perpendicular lines condition,the products of slope of both lines = - 1

Let The slope of other line = M'

So,  M' × m = - 1

Or, M' ×  [tex]\frac{-5}{2}[/tex] = - 1

Or, M' = [tex]\frac{-1}{\frac{-5}{2}}[/tex]

Or, M' =  [tex]\frac{2}{5}[/tex]

∴ The equation of line with slope M and passing through points ( -2 , 3) is

y = M' x + c'

Now , satisfying the points

So, 3 = [tex]\frac{2}{5}[/tex] × ( - 2 ) + c'

or, 3 =  [tex]\frac{- 4}{5}[/tex] + c'

Or, 3 × 5 = - 4 + 5× c'

∴  5 c' = 15 + 4

or, 5 c' = 19

Or, c' =  [tex]\frac{19}{5}[/tex]

So, The equation of line with slope  [tex]\frac{2}{5}[/tex]  and passing through points ( -2 , 3)

y =  [tex]\frac{2}{5}[/tex] x +  [tex]\frac{19}{5}[/tex]

y =  [tex]\frac{2 x + 19}{5}[/tex]

Or, 5 y = 2 x + 19

Or, 2 x - 5 y + 19 = 0

Hence The equation of line parallel to given line and passing through points ( - 2 , 3 ) is 5 x + 2 y + 4 = 0

And  The equation of line perpendicular to given line and passing through points ( - 2 , 3 ) is 2 x - 5 y + 19 = 0

Answer

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