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

$\huge\boxed{Answer}$

Let vertices are P , Q & R .

then P(-2,5) , Q(k,-4) & R(2k+1 , 10) .

We know that -

Area of triangle :-

= 1/2 {x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)}

= 1/2 {-2(-4 - 10) + k(10 - 5) + (2k+1)(5 + 4)}

= 1/2 { -2(-14) + 5k + 9(2k + 1) }

= 1/2 { 28 + 5k + 18k + 9 }

= 1/2 { 37 + 23k }

Area of triangle is already given as 53 unit² .

So ,

1/2 (37 + 23k) = 53

37 + 23k = 106

23k = 106 - 37

23k = 69

k = 3

Answer 2

Answer:

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Let vertices are P , Q & R .

then P(-2,5) , Q(k,-4) & R(2k+1 , 10) .

We know that -

Area of triangle :-

= 1/2 {x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)}

= 1/2 {-2(-4 - 10) + k(10 - 5) + (2k+1)(5 + 4)}

= 1/2 { -2(-14) + 5k + 9(2k + 1) }

= 1/2 { 28 + 5k + 18k + 9 }

= 1/2 { 37 + 23k }

Area of triangle is already given as 53 unit² .

So ,

1/2 (37 + 23k) = 53

37 + 23k = 106

23k = 106 - 37

23k = 69

k = 3

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