Question

Please Answer this question along with proper explanation~​

Answer 1

Given -

$\:\:\:\:\:\:\dashrightarrow\sf{a^2=(x_1-x_2)^2+(y_1-y_2)^2}$

$\:\:\:\:\:\:\dashrightarrow\sf{b^2=(x_2-x_3)^2+(y_2-y_3)^2}$

$\:\:\:\:\:\:\dashrightarrow\sf{c^2=(x_3-x_1)^2+(y_3-y_1)^2}$

$\:\:\:\:\:\:\dashrightarrow\sf{l \left|\begin{array}{ccc}x_1& y_1 & 1 \\ x_2 & y_2& 1 \\ x_3 & y_3& 1 \end{array} \right|^2=(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$

To find -

• the value of l.

Solution -

Let DEF be a triangle with sides a, b and c. Then we will apply the Heron's formula to find the area of triangle DEF,

Heron's formula -

$\: \: \: \: \: \: \: { \longrightarrow {\underline{\boxed{ \red{\bf{ Area\:of\:triangle=\sqrt{s(s - a)(s - b)(s - c) }}}}}}}$

Where,

$\green{\bf{s = \frac{a + b + c}{2}}}$

Further,

${\implies{\sf{(Area \: of \: triangle)^2= s(s - a)(s - b)(s - c)}}}$

${\implies{\sf{(Area \: of \: triangle)^2=\frac{a + b + c}{2} ( \frac{a + b + c}{2} - a)(\frac{a + b + c}{2}- b)(\frac{a + b + c}{2}- c)}}}$

${\implies{\sf{(Area \: of \: triangle)^2= (\frac{a + b + c}{2}) ( \frac{ b + c-a}{2} )(\frac{c + a -b}{2})(\frac{a + b -c}{2})}}}$

${\implies{\sf{(Area \: of \: triangle)^2= \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{16}}---- (i)}}$

As we know that the Area of any triangle DEF with vertices is given by,

$\implies\sf{Area\:of\:triangle=\frac{1}{2} \left|\begin{array}{ccc}x_1& y_1 & 1 \\ x_2 & y_2& 1 \\ x_3 & y_3& 1 \end{array} \right|}$

Squaring both sides,

$\implies\sf{(Area\:of\:triangle)^2=(\frac{1}{2})^2 \left|\begin{array}{ccc}x_1& y_1 & 1 \\ x_2 & y_2& 1 \\ x_3 & y_3& 1 \end{array} \right|^2}$

$\implies\sf{(Area\:of\:triangle)^2=\frac{1}{4} \left|\begin{array}{ccc}x_1& y_1 & 1 \\ x_2 & y_2& 1 \\ x_3 & y_3& 1 \end{array} \right|^2----(ii)}$

Equating (i) and (ii),

$\small{\implies\sf{\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{16} = \frac{1}{4} \left|\begin{array}{ccc}x_1& y_1 & 1 \\ x_2 & y_2& 1 \\ x_3 & y_3& 1 \end{array} \right|^2}}$

$\small{\implies\sf{(a+b+c)(b+c-a)(c+a-b)(a+b-c)= \frac{16}{4} \left|\begin{array}{ccc}x_1& y_1 & 1 \\ x_2 & y_2& 1 \\ x_3 & y_3& 1 \end{array} \right|^2}}$

$\implies\sf{l=\frac{\cancel{16}^4}{\cancel{4}_1}}$

$\implies\bf{l=4}$

Henceforth the value of l = 4

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Answer 2

Before moving to the question, let's recall for any triangle with vertices ${\bf{A(x_{1},y_{1})}}$, ${\bf{B(x_{2},y_{2})}}$ and ${\bf{C(x_{3},y_{3})}}$, the area of the triangle is just given by (1/2) times the determinant (not square), so we can assume here that the all x's and y's are the vertices of a triangle, do consider them as given in the attachment

If you use distance formula, then you can find that AB = a, BC = b and AC = c (By given information) also, now we also now another method to find area of a triangle if we know the semi perimeter which is half the sum of all sides, i.e here {(a+b+c)/2}, so if you use Heron's formula, we will be having :

${:\implies \quad \sf Area=\sqrt{\bigg(\dfrac{a+b+c}{2}\bigg)\bigg(\dfrac{a+b+c}{2}-a\bigg)\bigg(\dfrac{a+b+c}{2}-b\bigg)\bigg(\dfrac{a+b+c}{2}-c\bigg)}}$

${:\implies \quad \sf Area=\sqrt{\bigg(\dfrac{a+b+c}{2}\bigg)\bigg(\dfrac{b+c-a}{2}\bigg)\bigg(\dfrac{c+a-b}{2}\bigg)\bigg(\dfrac{a+b-c}{2}\bigg)}}$

Now, do square on both sides

${:\implies \quad \sf (Area)^{2}=\dfrac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{16}}$

Now, also consider

${:\implies \quad \sf Area=\dfrac{1}{2}\begin{vmatrix}x_{1}& y_{1}& 1\\ \\ x_{2}& y_{2}& 1\\ \\ x_{3}& y_{3}& 1\end{vmatrix}}$

Square both sides to obtain

${:\implies \quad \sf (Area)^{2}=\dfrac{1}{4}{\begin{vmatrix}x_{1}& y_{1}& 1\\ \\ x_{2}& y_{2}& 1\\ \\ x_{3}& y_{3}& 1\end{vmatrix}}^{2}}$

Now, we can thus obtain by using the formula we derived from Heron's Formula

${:\implies \quad \sf {\begin{vmatrix}x_{1}& y_{1}& 1\\ \\ x_{2}& y_{2}& 1\\ \\ x_{3}& y_{3}& 1\end{vmatrix}}^{2}=\sf \dfrac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4}}$

Now, substituting this value in the equation given

${:\implies \quad \bf I\sf \times \dfrac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4}=(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$

${:\implies \quad \boxed{\bf{I=4}}}$

Making 4 the required answer