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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)]
=0
Hence the points are collinear.


Directions: Solve the following problems.
Q 1: Show that (-5,1), (5,5) and (10,7) are collinear.
Answer:

Q 2: Prove that the points (4,2), (7,5) and (9,7) are collinear.
Answer:

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

Q 4: For what value of x will the points (x,-1), (2,1) and (4,5) lie on a line.
1
5
3

Q 5: Find the condition that the point (x,y) may lie on the line joining (3,4) and (-5,-6).
5x-4y+1 = 0
x-4y = 0
5x+1 = 0

Q 6: Find the value of k in order that the points (k,1), (5,5) and (10,7) may be collinear.
-5
7
8

Q 7: If (7,a),(-5,2) and (3,6) are collinear, find a.
12
8
15

Q 8: Show that the points (3,-2), (-1,1) and (-5,4) are collinear.
Answer:

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Question 10: This question is available to subscribers only!


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