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The vertices of a triangle ABC are $(λ, 2 - λ), (- λ + 1, 2 λ)$ and $(- 4 - λ, 6 - 2 λ)$ . If its area is 70 ${units}^{2}$ , find the number of integral values of $λ$ .

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1

2

4

10

answer is A.

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Detailed Solution

Given that, the vertices of

△

ABC are

A (λ, 2 - λ), B (- λ + 1, 2 λ), C (- 4 - λ, 6 - 2 λ)

.
Total area

= 70 uni t^{2}

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

are the coordinates of a triangle then,
Area

= \frac{1}{2} | (x_{1} y_{2} - x_{2} y_{1}) + (x_{2} y_{3} - x_{3} y_{2}) + (x_{3} y_{1} - x_{1} y_{3}) |

\Rightarrow 70 = \frac{1}{2} [(2 λ - 6 + 2 λ) + (1 - λ) [6 - 2 λ - 2 + λ] - (4 + λ)]

\Rightarrow λ (4 λ - 6) + (1 - λ) (4 - λ) - (4 + λ) (2 - 3 λ) = 140

\Rightarrow 4 λ^{2} - 6 λ + 4 - x - 4 λ + λ^{2} - [8 - 12 λ + 2 λ - 3 λ^{2}] = 140

\Rightarrow 5 λ^{2} - 10 λ + 4 - 8 + 12 λ - 2 λ + 3 λ^{2} = 140

\Rightarrow 8 λ^{2} + λ - 4 - 2 λ = 140

\Rightarrow 8 λ^{2} - λ - 136 = 0

General form of quadratic equation is

a x^{2} + bx + c = 0

,
Comparing with the general form,
a=8, b=-1, and c=-136.
The solution of the quadratic equation is given by,

x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

Substituting the values of a, b and c,

λ = \frac{- (- 1) \pm \sqrt{1^{2} - 4 (8) (- 136)}}{2 (8)}

\Rightarrow λ = \frac{1 \pm \sqrt{4353}}{16}

\Rightarrow λ = \frac{1 \pm 65.97}{16}

\Rightarrow λ = \frac{1 + 65.97}{16}

λ = \frac{1 - 65.97}{16}

\Rightarrow λ = 4.186

λ = - 4.061

Only valid value of

λ

4.186

since the negative value is ignored.

∴

Number of integral value of

λ

is 1.
Therefore, option (1) is correct.

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