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(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,
=
(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8)
=
Therefore, points A, B and C are collinear.
When −
k
+ 4 = − 4,
k
= 8.
When −
k
+ 4 = 4,
k
= 0.
Hence,
k
= 0, 8.
(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0,
k
) is given by the relation,
∴
k
− 4 = ± 4
When
k
− 4 = − 4,
k
= 0.
When
k
− 4 = 4,
k
= 8.
Hence,
k
= 0, 8.
Hence, the equation of the line joining the given points is
y
= 2
x
.
(ii) Let P (
x
,
y
) be any point on the line joining points A (3, 1) and
B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
Hence, the equation of the line joining the given points is
x
− 3
y
= 0.
It is given that the area of the triangle is ±35.
Therefore, we have:
⇒ 25 - 5k = ± 35
⇒ 5(5 - k) = ± 35
⇒ 5 - k = ± 7
When 5 −
k
= −7,
k
= 5 + 7 = 12.
When 5 −
k
= 7,
k
= 5 − 7 = −2.
Hence,
k
= 12, −2.
The correct answer is D.
Therefore, option (D) is correct.
