question is in attachment
Answers
The line segment P ( a,b ) and Q ( b,a ) and in it lies R ( x,y ) .
Hence the line is collinear .
The triangle formed between the 3 points will have an area = 0 .
1/2 | x₁ ( y₂ - y₃ ) + x₂ ( y₃ - y₁ ) + x₃ ( y₁ - y₂ ) | = 0
= | a ( a - y ) + b ( y - b ) + x ( b - a ) | = 0
= a² - ay + by - b² + bx - ax = 0
= bx + by - ay - ax + a² - b² = 0
= b ( x + y ) - a ( x + y ) + ( a + b )( a - b ) = 0
= ( x + y )( b - a ) - ( b - a )( a + b ) = 0
= ( b - a )( x + y - ( a + b ) ) = 0
=
Hence proved !
given-> line is collinear .
The triangle formed between the 3 points will have an area = 0 .
1/2 | x₁ ( y₂ - y₃ ) + x₂ ( y₃ - y₁ ) + x₃ ( y₁ - y₂ ) | = 0
= | a ( a - y ) + b ( y - b ) + x ( b - a ) | = 0
= bx + by - ay - ax + a² - b² = 0
= b ( x + y ) - a ( x + y ) + ( a + b )( a - b ) = 0
= ( x + y )( b - a ) - ( b - a )( a + b ) = 0
= ( b - a )( x + y - ( a + b ) ) = 0
= ( x + y ) = ( a + b ) .
hope it works ✌️
☺️❤️☺️❤️☺️❤️☺️❤️☺️❤️☺️