Question

THe area of triangle is 35 sq units with vertices (2, -6) , (5, 4) , and ( k , 4) then k is

Answer 1

Given:

• Area of a triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4)

To find:

• Value of k =?

Knowledge required:

• Area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is given by

$\purple{\bigstar}\boxed{\sf{Ar(\triangle)=\mid\dfrac{1}{2}\big[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big]\mid}}$

Solution:

Let,

• x₁ = 2 ; y₁ = -6
• x₂ = 5 ; y₂ = 4
• x₃ = k ; y₃ = 4

Then, Using formula for area of triangle

$\implies\sf{Ar(\triangle)=\mid\dfrac{1}{2}\big[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big]\mid}$

$\implies\sf{35=\dfrac{1}{2}\big[2(4-4)+5(4-(-6))+k(-6-4)\big]}$

$\implies\sf{35=\dfrac{1}{2}\big[2(0)+5(10)+k(-10)\big]}$

$\implies\sf{70=50-10k}$

$\implies\sf{10k=-20}$

$\implies\boxed{\boxed{\sf{k=-2}}}\;\;\;\purple{\bigstar}$

Therefore,

• value of k is -2.

Answer 2

$\overrightarrow{\underrightarrow{ \green{ \sf \ Given: }}}$

• Area of triangle = 35 sq. units
• Vertices of triangle are (2,-6), (5,4) and (k,4)

$\overrightarrow{\underrightarrow{ \green{ \sf \ Find: }}}$

• What will be the value of k

$\overrightarrow{\underrightarrow{ \green{ \sf \ Solution: }}}$

Now, we, know that

$\sf Area \: of \: triangle \: is \triangle = \frac{1}{2} \left|\begin{array}{ccc}x_{1} &y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array} \right|$

where,

• Area of triangle = 35 sq. units

Since area is always positive,

So, $\triangle$ can have positive and negative value

=> $\triangle$ = ± 35 sq. units

Also,

$\sf x_{1} = 2 , \qquad y_{1} = - 6\\ \sf x_{2} = 5, \qquad y_{2} = 4\\ \sf x_{3} =k, \qquad y_{3} = 4$

Putting values in area of Triangle

$\sf \implies ±35 = \frac{1}{2} \left|\begin{array}{ccc}2 & - 6&1\\5&4&1\\k&4&1\end{array} \right|$

$\sf \implies ±35 = \frac{1}{2} \bigg(2\left|\begin{array}{cc}4 & 1\\4&1\end{array} \right| - ( - 6)\left|\begin{array}{cc}5 & 1\\k&1\end{array} \right| + 1\left|\begin{array}{cc}5 & 4\\k&4\end{array} \right| \bigg)$

$\sf \implies ±35 = \frac{1}{2} (2(4 - 4) + 6(5 - k) + 1(20 - 4k))$

$\sf \implies ±35 = \frac{1}{2} (2(0) + 6(5 - k) + 1(20 - 4k))$

$\sf \implies ±35 = \frac{1}{2} (30 - 6k + 20 - 4k)$

$\sf \implies ±35 = \frac{1}{2} (50 - 10k)$

$\sf \implies ±35 \times 2 = (50 - 10k)$

$\sf \implies ±70= (50 - 10k)$

So, 70 = 50 - 10k or -70 = 50 - 10k

$\bold{ solving \: 70 = 50 - 10k} \\ \sf \to 70 - 50 = - 10k \\ \sf \to 20 = - 10k \\ \sf \to k = \frac{ 20}{ - 10} = - 2$

$\bold{ solving \: -70 = 50 - 10k} \\ \sf \to - 70 - 50 = - 10k \\ \sf \to - 120 = - 10k \\ \sf \to k = \frac{ - 120}{ - 10} = 12$

Hence, the required value of k is -2 or 12