# Co Ordinate Geometry - Quantitative Aptitude Problems and Solutions

Condition of collinearity of three ponits
```Three points A(x1, y1), B(x2, y2 and C(x3, y3) are collinear if and only if are ΔABC = 0.
A,B,C are colinear <=> x1(y2-y3) + x2(y3-y1) + x3(y1 - y2) = 0```

1. 21.   Find the value of k for which the points A(-2, 5), B(3, k) and C(6, 1) are collinear.

 A.) 5 B.) 4 C.) 7 D.) 1

x1 = -2, x2 = 3, x3 = 6 and y1 = 5, y2 = k, y3 = -1

Now Δ = 0 <=> -2(k + 1) + 3(-1 + 5) + 6(5 - k) = 0

<=> -2 (k+1) + 3(4) + 6(5-k) = 0

<=> -2k-2 + 12+30 -6k = 0

<=> 40 - 8k = 0

<=> -8k = -40

<=> k = 5.

2. 22.   Find the value of k for which the points A(-2, 6), B(5, k) and C(8, 3) are collinear.

 A.) 5 B.) 4 C.) 7 D.) 1

x1 = -2, x2 = 5, x3 = 8 and y1 = 6, y2 = k, y3 = 3

Now Δ = 0 <=> -2(k - 3) + 5(3 - 6) + 8(6 - k) = 0

<=> -2 (k-3) + 5(-3) + 8(6-k) = 0

<=> -2k + 6 - 15 + 48 - 8k = 0

<=> 39 - 10k = 0

<=> k = -(39/10)

<=> k = 3.9.

3. 23.  If the A(2, 3), B(5, k), and C(6, 7) are collinear, then k = ?

 A.) 11 B.) 12 C.) 18 D.) 6

x1 = 2, x2 = 5, x3 = 6 and y1 = 3, y2 = k, y3 = 7

Now Δ = 0 <=> 2(k - 7) + 5(7 - 3) + 6(3 - k) = 0

<=> 1/2 [2 (k-7) + 5(4) + 6(3-k)] = 0

<=> 2k - 14 + 20 + 18 - 6k = 0

<=> 24 - 4k = 0

<=> 4k = 24

<=> k = 6.

4. 24.  If the points A(1, 2), B(2, 4), and C(k, 6) are collinear, then k = ?

 A.) 3 B.) -3 C.) 4 D.) -4

x1 = 1, x2 = 2, x3 = k and y1 = 2, y2 = 4, y3 = 6

= 1(4 - 6) + 2 (6 - 2) + k(2 - 4)

= -2 + 12 - 4 + 2k - 4k = 0

= 6 - 2k = 0

= -2k = 6

= k = -3.

Quantitative Aptitude Topics

Quantitative Aptitude Topics

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