Q.51 If the area of a triangle with the vertices (k, 0), (2, 0) and (0, −2) is 2 square units, the value of k is _________.

Q.51 If the area of a triangle with the vertices (k, 0), (2, 0) and (0, −2)
is 2 square units, the value of k is _________.

 

The area of the triangle formed by vertices (k, 0), (2, 0), and (0, -2) equals 2 square units when k = 0 or k = 4, derived using the coordinate area formula.

Area Formula

The area A of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is:

A = 1/2 | x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2) |

This determinant-based shoelace formula computes signed area, with absolute value making the result positive.

Step-by-Step Solution

Using the points:

  • (x1, y1) = (k, 0)
  • (x2, y2) = (2, 0)
  • (x3, y3) = (0, −2)

Substitute into the formula:

A = 1/2 | k(0 − (−2)) + 2((−2) − 0) + 0(0 − 0) |
  = 1/2 | 2k − 4 |

Set A = 2:

1/2 |2k − 4| = 2
⇒ |2k − 4| = 4
⇒ |k − 2| = 2

Thus:

k − 2 = 2  ⇒  k = 4  
k − 2 = −2 ⇒  k = 0

Possible Values Explained

  • k = 0: Triangle with base 2 and height 2 gives area = 2.
  • k = 4: Substituting gives 1/2 |2(4) − 4| = 2.

Both satisfy the required area. Some exams expect k=4, but k=0 is equally valid mathematically.

Verification Table

k Value Base on x-axis Height Computed Area Matches 2?
0 2 units 2 2 Yes
4 2 units (effective) 2 2 Yes

 

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