Question

if the area of the triangle formed by the points (2a,-3a), (3, -2) and (5,-2) is 6 sq units, then
the value of a is:
(a) 2
(6) -2
(d) -1
(c) 1​

Answer 1

Step-by-step explanation:

Answer: The required value of x is \dfrac{14}{3},\dfrac{5}{3}.

3

14

,

3

5

.

Step-by-step explanation: Given that the area of the triangle formed by the vertices \left(x,\frac{4}{3}\right),(x,

3

4

), (-2, 6) and (3, 1) is 5 square units.

We are to find the value of x.

We know that

the area of a triangle formed by the vertices (a, b), (c, d) and (e, f) is given by

A=|\dfrac{1}{2}\{a(d-f)+c(f-b)+e(b-d)\}|.A=∣

2

1

{a(d−f)+c(f−b)+e(b−d)}∣.

So, according to the given information, we have

\begin{lgathered}5=|\dfrac{1}{2}\{x(6-1)-2(1-\frac{4}{3})+3(\frac{4}{3}-6)\}|\\\\\\\Rightarrow 5=|\dfrac{1}{2}(5x+\frac{2}{3}-14)|\\\\\\\Rightarrow 5=|\dfrac{1}{2}(5x-\frac{40}{3})|\\\\\\\Rightarrow 30=|6x-40|\\\\\Rightarrow 6x-40=30,~~~~6x-40=-30\\\\\Rightarrow 6x=70,~~~~\Rightarrow 6x=10\\\\\Rightarrow x=\dfrac{14}{3},\dfrac{5}{3}.\end{lgathered}

5=∣

2

1

{x(6−1)−2(1−

3

4

)+3(

3

4

−6)}∣

⇒5=∣

2

1

(5x+

3

2

−14)∣

⇒5=∣

2

1

(5x−

3

40

)∣

⇒30=∣6x−40∣

⇒6x−40=30, 6x−40=−30

⇒6x=70, ⇒6x=10

⇒x=

3

14

,

3

5

.

Thus, the required value of x is \dfrac{14}{3},\dfrac{5}{3}.

3

14

,

3

5

.

Answer 2

The value of a=-4/3 or a=8/3

Step-by-step explanation:

Let A=(2a,-3a)

B=(3,-2)

C=(5,-2)

Area of triangle=6 square units

We know that

Area of triangle=$\frac{1}{2}\mid{x_1(y_2-y_3)+x_2(y_3-y_1)_+x_3(y_1-y_2)}\mid$

Using the formula

$6=\frac{1}{2}\mid{2a(-2+2)+3(-2+3a)+5(-3a+2)}\mid$

$12=\mid{-6+9a-15a+10}\mid$

$12=\mid{4-6a}\mid$

$4-6a=\pm 12$

$4-6a=12$

$12-4=-6a$

$6a=-8$

$a=-\frac{8}{6}=-\frac{4}{3}$

$4-6a=-12$

$4+12=6a$

$6a=16$

$a=\frac{16}{6}=\frac{8}{3}$

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