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High School Mathematics - 2
9.21 Collinearity of Points

If three points (x1, y1), (x2, y2), (x3, y3) lie on a straight line, the area of the triangle formed by them is zero.
i.e1/2(x1y2-x2y1+x2y3-x3y2+x3y1-x1y3) = 0

Example: Show that the points (a,b+c),(b,c+a) and (c, a+b) are collinear.
The area of the triangle formed by the three points is
= 1/2[a(c+a)-b(b+c)+b(a+b)-c(c+a)+c(b+c)-a(a+b)]
Hence the points are collinear.

Directions: Solve the following problems.
Q 1: For what value of x will the points (x,-1), (2,1) and (4,5) lie on a line.

Q 2: If (7,a),(-5,2) and (3,6) are collinear, find a.

Q 3: Show that the points (1,4), (3,-2) and (-3,16) are collinear.

Q 4: Find the value of k in order that the points (k,1), (5,5) and (10,7) may be collinear.

Question 5: This question is available to subscribers only!

Question 6: This question is available to subscribers only!

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