Question

QUESTION:
The four vertices of a quadrilateral are (1, 2), (−5, 6), (7, −4) and (k, −2) taken in order. If the area of the quadrilateral is zero, find the value of k.

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Answer 1

Answer:

Formula:

Area of quadrilateral = area of triangle 1 + area of triangle 2

Area of triangle =  

2

1

​  

[x  

1

​  

(y  

2

​  

−y  

3

​  

)+x  

2

​  

(y  

3

​  

−y  

1

​  

)+x  

3

​  

(y  

1

​  

−y  

2

​  

)]

Given,

A(1,2),B(−5,6),C(7,−4),D(k,−2)

Let the first triangle be, ABC

Area of  △ABC=  

2

1

​  

[1(6−(−4))−5(−4−2)+7(2−6)]

=  

2

1

​  

[10+30−28]

=6sq.units

Area of  △ACD=  

2

1

​  

[1(−4+2)+7(−2−2)+k(2+4)]

=  

2

1

​  

[−2−28+6k]

=−15+3ksq.units

Area of quadrilateral =6−15+3k=0

k=3

Step-by-step explanation:

hope it the the right answer...

Answer 2

$\bold{\large{\boxed{\red{\tt{ANSWER}}}}}$

☆ GIVEN :-

The four vertices of a quadrilateral are (1, 2), (−5, 6), (7, −4) and (k, −2)

And also Area of quadrilateral is 0

☆ TO FIND :-

Value of k

☆ SOLUTION :-

The four vertices of a quadrilateral are A(1, 2), B(−5, 6), C(7, −4) and D(k, −2)

We know area of triangle formed by three points ( x1 , y1 ) , ( x2 , y2 ) and ( x3 , y3 ) is given by ,

$Area \: of \:Triangle = \pink{\tt{ \frac{1}{2} [ x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 )]}}$

Now ,

Area of Triangle ABC ,

Where taking three points A(1, 2), B(−5, 6), C(7, -4)

$Area \: of \: Triangle \: ABC = \frac{1}{2} [ 1 ( 6-(-4) ) + (-5) ( -4-2 ) + 7 ( 2-6 ) ]$

$\rightarrow{ \frac{1}{2} [ 1 ( 6+4 ) -5 ( -4-2 ) + 7 ( 2-6 ) ]}$

$\rightarrow{ \frac{1}{2} [ 6 + 4 + 20 + 10 + 14 - 42 ]}$

$\rightarrow{ \frac{1}{2} [ 54 - 42 ]}$

$\rightarrow{ \frac{1}{2} [ 12 ]}$

$\rightarrow{6 \: sq.units}$

Now ,

Area of Triangle ACD ,

Where taking three points A(1 , 2) , C(7, -4) ,D(k, -2)

$Area \: of \: Triangle \: ACD = \frac{1}{2} [ 1( -4-2 ) + 7( -2-2 ) + k( 2+4 ) ]$

$\rightarrow{ \frac{1}{2} [ -4 -2 -14 -14 +2k +4k ]}$

$\rightarrow{ \frac{1}{2} [ -30 + 6k ]}$

$\rightarrow [ 3k - 15 ] sq.units$

$\purple{Area ( ABCD ) = Area ( ABC ) + Area ( ACD )}$

$\rightarrow{0 = 6 + 3k - 15}$

$\rightarrow{0 = 6 - 15 + 3k}$

$\rightarrow{0 = - 9 + 3k}$

$\rightarrow{9 = 3k}$

$\rightarrow{k = \frac{9}{3} }$

$\rightarrow{k = 3}$

$\bold{\large{\blue{\tt{I \: hope \: it \: helps \: u \: Utkarsh \: !!}}}}$