Question

COORDINATE GEOMETRY
CLASS 10
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•Please answer all the given question in the picture.
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THANKS...​

Answer 1

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➤find the area of a triangle abc with a(1 -4) and midpoints of sides through A being (2,-1) and (0,-1)

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Answer:

$12 \:sq.\:units$

REFER TO THE ATTACHMENT FOR DIAGRAM..

☆ Coordinates of D :

D(2,-1) = D[(a+1)/2 + (b-4)/2]

By comparing coordinates:

✪ 2 = (a+1)/2

✪2*2 = a+1

✪4 = a+1

✪a = 3

✪-1 = (b-4)/2

✪-1*2 = b-4

✪-2 = b-4

✪b = 2

Therefore, coordinates of B are B(3,2).

☆ Coordinates of E :

E(0,-1) = E[(c+1)/2 + (d-4)/2]

By comparing coordinates :

$0 = (c+1)/2$

$2*0 = c+1$

$c+1 = 0$

$c = -1$

$-1 = (d-4)/2$

$-1*2 = d-4$

$-2 = d-4$

$d = 2$

Therefore, coordinates of C are C(-1,2).

☆ In triangle ABC:

A(1,-4)

B(3,2)

C(-1,2)

☆ Let :-

x1 = 1

y1 = -4

x2 = 3

y2 = 2

x3 = -1

y3 = 2

☆ Area of triangle ABC =

1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

= 1/2[1(2-2)+3(2+4)-1(-4-2)]

= 1/2[1(2-2)+3(2+4)-1(-4-2)]

= 1/2[1(0)+3(6)-1(-6)]

= 1/2(0+18+6)

= 1/2(24)

$\large\fbox{\color{red}{ = 12 sq.units}}$

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Answer 2

Answer:

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REFER TO THE ATTACHMENT FOR DIAGRAM..

☆ Coordinates of D :

D(2,-1) = D[(a+1)/2 + (b-4)/2]

By comparing coordinates:

✪ 2 = (a+1)/2

✪2*2 = a+1

✪4 = a+1

✪a = 3

✪-1 = (b-4)/2

✪-1*2 = b-4

✪-2 = b-4

✪b = 2

Therefore, coordinates of B are B(3,2).

☆ Coordinates of E :

E(0,-1) = E[(c+1)/2 + (d-4)/2]

By comparing coordinates :

0 = (c+1)/20=(c+1)/2

2*0 = c+12∗0=c+1

c+1 = 0c+1=0

c = -1. ,c = − 1

-1 = (d-4)/2−1=(d−4)/2

-1*2 = d-4−1∗2=d−4

-2 = d-4−2=d−4

d = 2d=2

Therefore, coordinates of C are C(-1,2).

☆ In triangle ABC:

A(1,-4)

B(3,2)

C(-1,2)

☆ Let :-

x1 = 1

y1 = -4

x2 = 3

y2 = 2

x3 = -1

y3 = 2

☆ Area of triangle ABC =

1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

= 1/2[1(2-2)+3(2+4)-1(-4-2)]

= 1/2[1(2-2)+3(2+4)-1(-4-2)]

= 1/2[1(0)+3(6)-1(-6)]

= 1/2(0+18+6)

= 1/2(24)

= 12 sq.units

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