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Svet_ta [14]
3 years ago
7

PLEASE HELP!! I NEED THIS TO PASS!! WILL MARK BRANLIEST AND GIVE 30 POINTS!! THANKS!

Mathematics
1 answer:
Vesnalui [34]3 years ago
8 0

Answer:

Theorem : Opposite sides of a parallelogram are congruent or equal.

Let us suppose a parallelogram ABCD.  

Given:AB\parallel CD and BC\parallel AD (According to the definition of parallelogram)

We have to prove that: AB is congruent to CD and BC is congruent to AD.

Prove: let us take two triangles, \bigtriangleup ACD and\bigtriangleup ABC

In these two triangles, \angle1=\angle2 { By the definition of alternative interior angles}

Similarly,                        \angle4=\angle3

And,                                AC=AC (common segment)

By  ASA, \bigtriangleup ACD \cong \bigtriangleup ABC

thus By the property of congruent triangle, we can say that corresponding  sides of \bigtriangleup ACD and \bigtriangleup ABC are also congruent.

Thus, AB is congruent to CD and BC is congruent to AD.

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Determine the gradient of the straight line 2x-3y+9=0. Find the equation of the straight line through the origin which is perpen
frutty [35]

Answer: 3

x

−

2

y

−

15

=

0

Explanation:

We know that,

the slope of the line  

a

x

+

b

y

+

c

=

0

is  

m

=

−

a

b

∴

The slope of the line  

2

x

+

3

y

=

9

is  

m

1

=

−

2

3

∴

The slope of the line perpendicular to  

2

x

+

3

y

=

9

is  

m

2

=

−

1

m

1

=

−

1

−

2

3

=

3

2

.

Hence,the equn.of line passing through  

(

3

,

−

3

)

and

m

2

=

3

2

is

y

−

(

−

3

)

=

3

2

(

x

−

3

)

y

+

3

=

3

2

(

x

−

3

)

⇒

2

y

+

6

=

3

x

−

9

⇒

3

x

−

2

y

−

15

=

0

Note:

The equn.of line passing through  

A

(

x

1

,

y

1

)

and

with slope

m

is

y

−

y

1

=

m

(

x

−

x

1

)3

x

−

2

y

−

15

=

0

Explanation:

We know that,

the slope of the line  

a

x

+

b

y

+

c

=

0

is  

m

=

−

a

b

∴

The slope of the line  

2

x

+

3

y

=

9

is  

m

1

=

−

2

3

∴

The slope of the line perpendicular to  

2

x

+

3

y

=

9

is  

m

2

=

−

1

m

1

=

−

1

−

2

3

=

3

2

.

Hence,the equn.of line passing through  

(

3

,

−

3

)

and

m

2

=

3

2

is

y

−

(

−

3

)

=

3

2

(

x

−

3

)

y

+

3

=

3

2

(

x

−

3

)

⇒

2

y

+

6

=

3

x

−

9

⇒

3

x

−

2

y

−

15

=

0

Note:

The equn.of line passing through  

A

(

x

1

,

y

1

)

and

with slope

m

is

y

−

y

1

=

m

(

x

−

Explanation:

the equation of a line in  

slope-intercept form

is.

∙

x

y

=

m

x

+

b

where m is the slope and b the y-intercept

rearrange  

2

x

+

3

y

=

9

into this form

⇒

3

y

=

−

2

x

+

9

⇒

y

=

−

2

3

x

+

3

←

in slope-intercept form

with slope m  

=

−

2

3

Given a line with slope then the slope of a line

perpendicular to it is

∙

x

m

perpendicular

=

−

1

m

⇒

m

perpendicular

=

−

1

−

2

3

=

3

2

⇒

y

=

3

2

x

+

b

←

is the partial equation

to find b substitute  

(

3

,

−

3

)

into the partial equation

−

3

=

9

2

+

b

⇒

b

=

−

6

2

−

9

2

=

−

15

2

⇒

y

=

3

2

x

−

15

2

←

equation of perpendicular lineExplanation:

the equation of a line in  

slope-intercept form

is.

∙

x

y

=

m

x

+

b

where m is the slope and b the y-intercept

rearrange  

2

x

+

3

y

=

9

into this form

⇒

3

y

=

−

2

x

+

9

⇒

y

=

−

2

3

x

+

3

←

in slope-intercept form

with slope m  

=

−

2

3

Given a line with slope then the slope of a line

perpendicular to it is

∙

x

m

perpendicular

=

−

1

m

⇒

m

perpendicular

=

−

1

−

2

3

=

3

2

⇒

y

=

3

2

x

+

b

←

is the partial equation

to find b substitute  

(

3

,

−

3

)

into the partial equation

−

3

=

9

2

+

b

⇒

b

=

−

6

2

−

9

2

=

−

15

2

⇒

y

=

3

2

x

−

15

2

←

equation of perpendicular line

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