Question 7 of 58intermediate🔧 ApplyLong Answer2 marks

Show that points A (a,  b+c)A\,(a,\; b + c), B (b,  c+a)B\,(b,\; c + a), C (c,  a+b)C\,(c,\; a + b) are collinear.

Correct Answer

To show that points A(a,  b+c)A(a,\; b+c), B(b,  c+a)B(b,\; c+a), C(c,  a+b)C(c,\; a+b) are collinear, we need to prove that the area of triangle ABCABC equals zero.

Step 1: Set up the determinant for area of triangle

The area of triangle with vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), (x3,y3)(x_3, y_3) is given by:

Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

Step 2: Substitute the coordinates

For points A(a,  b+c)A(a,\; b+c), B(b,  c+a)B(b,\; c+a), C(c,  a+b)C(c,\; a+b):

Area=12∣a[(c+a)−(a+b)]+b[(a+b)−(b+c)]+c[(b+c)−(c+a)]∣\text{Area} = \frac{1}{2} \left| a[(c+a) - (a+b)] + b[(a+b) - (b+c)] + c[(b+c) - (c+a)] \right|

Step 3: Simplify the expression

=12∣a[c−b]+b[a−c]+c[b−a]∣= \frac{1}{2} \left| a[c - b] + b[a - c] + c[b - a] \right| =12∣ac−ab+ab−bc+bc−ac∣= \frac{1}{2} \left| ac - ab + ab - bc + bc - ac \right| =12∣0∣= \frac{1}{2} \left| 0 \right| =0= 0

Since the area of triangle ABCABC is zero, the three points AA, BB, and CC lie on the same straight line. Hence, the points are collinear.

Exercise: EXERCISE 4.2 | Q: 2 | (Page 8)
For More Understanding

Explanation

This question tests the concept that three points are collinear if and only if the area of the triangle formed by them is zero. The textbook context shows this method in Exercise 4.2, where the area of a triangle is calculated using determinants. When the determinant equals zero, the area becomes zero, confirming collinearity.

Solution Steps

  1. Step 1: Recall that three points are collinear if area of triangle = 0

  2. Step 2: Apply the area formula using coordinates of A, B, C

  3. Step 3: Simplify the expression to show it equals zero

  4. Step 4: Conclude that points are collinear since area = 0

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