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### High School Mathematics - 29.21 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: For what value of x will the points (x,-1), (2,1) and (4,5) lie on a line.315 Q 2: Find the condition that the point (x,y) may lie on the line joining (3,4) and (-5,-6).5x-4y+1 = 0x-4y = 05x+1 = 0 Q 3: Show that the points (1,4), (3,-2) and (-3,16) are collinear.Answer: Q 4: If (7,a),(-5,2) and (3,6) are collinear, find a.15128 Question 5: This question is available to subscribers only! Question 6: This question is available to subscribers only!