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Nathaliasmiles
13.12.2019 •
Mathematics
Find the equation of the lines parallel and perpendicular to the line 5x+2y=12 through the point (-2,3)
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Ответ:
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 =
x + ![\frac{12}{2}](/tpl/images/0417/1959/f3506.png)
Or, y =
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 =![\frac{-5}{2}](/tpl/images/0417/1959/05885.png)
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 =![\frac{-5}{2}](/tpl/images/0417/1959/05885.png)
∴ The equation of line with slope M and passing through points ( -2 , 3) is
y = M x + c
Now , satisfying the points
So, 3 =
× ( - 2 ) + c
or, 3 =
+ c
Or, 3 = 5 + c
∴ c = 3 - 5 = - 2
c = - 2
So, The equation of line with slope
and passing through points ( -2 , 3)
y =
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' ×
= - 1
Or, M' =![\frac{-1}{\frac{-5}{2}}](/tpl/images/0417/1959/68131.png)
Or, M' =![\frac{2}{5}](/tpl/images/0417/1959/c236b.png)
∴ The equation of line with slope M and passing through points ( -2 , 3) is
y = M' x + c'
Now , satisfying the points
So, 3 =
× ( - 2 ) + c'
or, 3 =
+ c'
Or, 3 × 5 = - 4 + 5× c'
∴ 5 c' = 15 + 4
or, 5 c' = 19
Or, c' =![\frac{19}{5}](/tpl/images/0417/1959/41c86.png)
So, The equation of line with slope
and passing through points ( -2 , 3)
y =
x + ![\frac{19}{5}](/tpl/images/0417/1959/41c86.png)
y =![\frac{2 x + 19}{5}](/tpl/images/0417/1959/4eae4.png)
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
Ответ:
2x+ 2y = -2
Step-by-step explanation:
Given the equation of a line as x + y = -1
To get the equation of the line P so the set of equations has infinitely many solutions, we will have to multiply the given equation by a factor
Multiplying the given equation by a factor of 2;
Given x + y = -1
Multiplying both sides by 2;
2(x + y) = 2(-1)
2x+ 2y = -2
Hence the required equation is 2x+ 2y = -2