mariam1970
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Does anyone get this ? if so can u lmk the answer thxxxx:)

Does Anyone Get This If So Can U Lmk The Answer Thxxxx class=

Sagot :

Given:

In parallelogram ABCD, two of its vertices are A(-4,0) and B(0,3).

To find:

The equation that represents a line that contain CD.

Solution:

We have,

A(-4,0) and B(0,3)

Slope of AB is

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

[tex]m=\dfrac{3-0}{0-(-4)}[/tex]

[tex]m=\dfrac{3}{4}[/tex]

The slope of line AB is [tex]\dfrac{3}{4}[/tex].

Opposite sides of a parallelogram are parallel and slopes of parallel lines are equal.

In parallelogram ABCD, AB and CD are opposite sides. So, their slopes must be equal.

Slope of line AB = Slope of line CD = [tex]\dfrac{3}{4}[/tex]

The slope intercept form of a line is

[tex]y=mx+b[/tex]

Where, m is slope and b is y-intercept.

Slope of line CD is [tex]\dfrac{3}{4}[/tex], it means the line must be of the form

[tex]y=\dfrac{3}{4}x+b[/tex]

Coefficient of x is [tex]\dfrac{3}{4}[/tex] only in option a.

Therefore, the correct option is a.

Answer:

a)   [tex]y = \frac{3}{4} x + 3[/tex]

The equation of the straight line CD is    [tex]y = \frac{3}{4} x + 3[/tex]

Step-by-step explanation:

Step(i):-

Given points are A (-4,0) and B( 0,3)

Slope of the line

   [tex]m = \frac{y_{2} - y_{1} }{x_{2} - x_{1} } = \frac{3-0}{0-(-4)} = \frac{3}{4}[/tex]

Slope of the line     [tex]m = \frac{3}{4}[/tex]

Step(ii):-

The equation of the straight line passing through the point ( -4,0) and having slope    [tex]m = \frac{3}{4}[/tex]

y - y₁ = m(x-x₁)

[tex]y -0 = \frac{3}{4} ( x- (-4))[/tex]

[tex]y = \frac{3}{4} x + 3[/tex]

The equation of the straight line AB is    [tex]y = \frac{3}{4} x + 3[/tex]

Step(iii):-

CD is parallel to the line AB

The equation of the straight line AB is    [tex]y = \frac{3}{4} x + 3[/tex]

                                     4 y = 3 x + 12

                                    3x - 4y +12 =0

The equation of the Parallel line is   3 x -4y +k=0

Passes through the point ( 0,3)

                                 -12 +k=0

                                      k =12

The equation of the Parallel line is   3 x -4y +12=0

                                                       4 y = 3 x+12

The equation of the straight line CD is    [tex]y = \frac{3}{4} x + 3[/tex]

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