Question

1.If (k, 2), (2, 4), (3, 2) are the vertices of atriangle of area 4sq.units, then the value of k is write in solution​

Answer 1

Answer :-

→ k = 7

To Find :-

Value of k

Used Formula :-

$\triangle = \frac{1}{2} \bigg\{\sf{x_{1}(y_{2}-y_{3}) + x_{2}(y_{3} - y_{1} ) + x_{3}(y_{1} - y_{2}) \: } \bigg\}$

Explanation :-

Let given points are A(k,2) ,B(2,4) and C(3,2)

$\star \sf{ \: A(k,2) \to x_{1} =k , y_{1} =2} \\ \\ \star \sf{ \: B(2,4) \to x_{2} =2 , y_{2} = 4}\\ \\ \star \sf{ \: C(3,2) \to \: x_{3} = 3, y_{3} = 2 \: }$

Also given that area of triangle formed by the given vertices is 4 square units .

Hence ,apply the above values in the given formula -

$\to \sf{ 4 = \frac{1}{2} \bigg\{ k(4 - 2) + 2 (2 - 2)+ 3(2 - 4) \bigg \}} \\ \\ \to \sf{ 8 = 2k \: - 6} \\ \: \\ \to \sf{2k = 8 + 6} \: \\ \: \\ \to \sf{ 2k = 14} \\ \\ \to \boxed{\sf{k = 7 \: }}$

Hence value of k is 7 ,since Point A is (7,2)

Answer 2

Given:

A ΔABC with coordinate of vertices A(k,2), B(2,4) and C(3,2)

Ar(ΔABC)= 4 sq units

To Find:

• Value of k

Solution:

We know that,

$\bf{Area\:of\:\triangle=\Bigg|\dfrac{1}{2}\Big(x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\Big)\Bigg|}$

Let

• A(k,2)⇒(x₁,y₁)
• B(2,4)⇒(x₂,y₂)
• C(3,2)⇒(x₃,y₃)

Here,

x₁=k, x₂=2, x₃=3

y₁=2, y₂=4, y₃=2

So,

$\sf{Area\:of\:\triangle ABC=\dfrac{1}{2}\Bigg|\Big(k(4-2)+2(2-2)+3(2-4)\Big)\Bigg|}$

$\sf{4=\dfrac{1}{2}\Bigg|\Big(k(2)+2(0)+3(-2)\Big)\Bigg|}$

$\sf{\dfrac{1}{2}\Bigg|\Big(2k+0-6\Big)\Bigg|=4}$

$\sf{\Bigg|\Big(2k-6\Big)\Bigg|=4\times2}$

$\sf{2k-6=8}$

$\sf{2k=8+6}$

$\sf{2k=14}$

$\sf{k=\dfrac{\cancel{14}}{\cancel{2}}}$

$\sf{k=7}$

Hence, the value of k is 7.