For what value of P are the points (2,1),(p,-1) and (-1,3) collinear?

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Uploaded: 2023-05-17T14:00:00.000Z
Duration: 20 min 32 s
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A
$4$
B
5
C
$8$
D
$9$

Solution:

Given points are

(2, 1), (p, - 1) and (- 1, 3)

.
If area of

Δ ABC

is zero, then points A, B, and C are collinear.
Let

A (2, 1), B (p, - 1) and C (- 1, 3)

.
Since the points are collinear, the area must be zero.
So,

Area of ΔABC = 0

\Rightarrow \frac{1}{2} |[x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]| = 0

\Rightarrow \frac{1}{2} |[2 (- 1 - 3) + p (3 - 1) + (- 1) (1 - (- 1))]| = 0

\Rightarrow \frac{1}{2} |[2 (- 4) + 2 p - 2]| = 0

Solving further,

\Rightarrow \frac{1}{2} [2 p - 10] = 0

\Rightarrow p - 5 = 0

\Rightarrow p = 5

Thus, the points

A (2, 1), B (p, - 1) a nd C (- 1, 3)

are collinear for

p = 5

.
Hence, option 2 is correct.

For what value of P are the points $(2, 1), (p, - 1) and (- 1, 3)$ collinear?

Solution:

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