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3 years ago

Solve for x ~

"x {}^{2} - 5x + 6 = 0" alt="x {}^{2} - 5x + 6 = 0" align="absmiddle" class="latex-formula">

thankyou ~

Mathematics

1 answer:

lubasha [3.4K]3 years ago

6 0

$x^2 -5x +6 =0\\\\\implies x^2 -3x -2x +6 = 0\\\\\implies x(x-3) -2(x-3) =0\\\\\implies (x-3)(x-2)=0\\\\\implies x =3, ~ x = 2$

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

muminat

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 = $\frac{-5}{2}$ x + $\frac{12}{2}$

Or, y = $\frac{-5}{2}$ 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 = $\frac{-5}{2}$

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 = $\frac{-5}{2}$

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

y = M x + c

Now , satisfying the points

So, 3 = $\frac{-5}{2}$ × ( - 2 ) + c

or, 3 = $\frac{10}{2}$ + c

Or, 3 = 5 + c

∴ c = 3 - 5 = - 2

c = - 2

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

y = $\frac{-5}{2}$ x - 2

or, 2 y = - 5 x - 4

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

<u>Similarly</u>

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' × $\frac{-5}{2}$ = - 1

Or, M' = $\frac{-1}{\frac{-5}{2}}$

Or, M' = $\frac{2}{5}$

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

y = M' x + c'

Now , satisfying the points

So, 3 = $\frac{2}{5}$ × ( - 2 ) + c'

or, 3 = $\frac{- 4}{5}$ + c'

Or, 3 × 5 = - 4 + 5× c'

∴ 5 c' = 15 + 4

or, 5 c' = 19

Or, c' = $\frac{19}{5}$

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

y = $\frac{2}{5}$ x + $\frac{19}{5}$

y = $\frac{2 x + 19}{5}$

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

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