हिंदी

Write the condition of collinearity of points (x1, y1), (x2, y2) and (x3, y3).

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प्रश्न

Write the condition of collinearity of points (x1, y1), (x2, y2) and (x3, y3).

 
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उत्तर

The condition for co linearity of three points` (x_1 , y_1) , (x_2 , y_2) " and  " (x_3 , y_3 )` is that the area enclosed by them should be equal to 0.

The formula for the area ‘A’ encompassed by three points ` (x_1 , y_1) , (x_2 , y_2) " and  " (x_3 , y_3 )`   and  is given by the formula,

`A = 1/2 |[x_1-x_2   y_1 - y_2],[x_2 - x_3    y_2 - y_3 ]|`

`A = 1/2 | (x_2 - x_2 )(y_2 - y_3 ) -(x_2 - x_3 )(y_1 - y_2 )|`

Thus for the three points to be collinear we need to have,

`1/2 | (x_1 - x_2 )( y_2 - y_3 )- (x_2 - x_3 )(y_1 -y_2)|=0`

    `|(x_1 -x_2)(y_2 - y_3)-(x_2 - x_3)(y_1 - y_2)|=0`

The area ‘A’ encompassed by three points `(x_1 , y_1) , (x_2 , y_2) " and  " (x_3 , y_3 )`   ,  is also given by the formula,

`A = 1/2 |x_1 (y_2 - y_3 ) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2 )|`

Thus for the three points to be collinear we can also have,

` 1/2 |x_1 (y_2 - y_3 ) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2 )|`= 0

    `x_1 (y_2 - y_3)+x_2 (y_3-y_1)+x_3(y_1 - y_2) = 0`

 

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अध्याय 6: Co-ordinate Geometry - Exercise 6.6 [पृष्ठ ६२]

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आर.डी. शर्मा Mathematics [English] Class 10
अध्याय 6 Co-ordinate Geometry
Exercise 6.6 | Q 17 | पृष्ठ ६२

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