Question

The value of p, for which the points A(3,1), B(5,p) and C(7.-5) are collinear, is​

Answer 1

Answer:

Step-by-step explanation:

For Collinear the area of triangle is 0.

so,

ar. of Triangle = 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]

0 = 1/2 [ 3(5+3) + 5(-5-1) + 7(1-p) ]

0 = 1/2 ( 24 - 30 + 7 -7p )

0 = 1/2 (1 - 7p)

0 = 1 - 7p

7p = 1

p = 1/7

Therefore, for collinear points the value of "p" is 1/7.

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Answer 2

The value of p is - 2.

Given: The points A ( 3, 1), B ( 5, p ), and C ( 7, -5 ).

To Find: The value of p.

Solution:

• We know that the collinearity of 3 points means they lie on the same line. So, we can assume that if they form a triangle then the area of that triangle must be zero.
• The formula for the area of a triangle using 3 points is given by,

 Area of triangle = 1/2 × [ x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 ) ]     ....(1)

Where the points are ( x1, y1 ), ( x2, y2 ), and ( x3, y3 ).

Coming to the numerical, we are given;

A ( 3, 1), B ( 5, p ), and C ( 7, -5 )

Putting respective values in (1), we get;

   Area of triangle = 1/2 × [ 3 ( 5 + p ) + 5 ( -5 - 1 ) + 7 ( 1 - p ) ]    

⇒    0   =  15 + 3p - 30 + 7 - 7p

⇒    4p = - 8

⇒     p   = - 2

Hence, the value of p is - 2.

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