Question

If the point (3,-4) (1,6) and (-2,3) are the vertices of a triangle, find the area ​

Answer 1

Given:

• Points (3,-4), (1,6) and (-2,3) are the vertices of a triangle.

To find:

• Area of the triangle.

Solution:

To find the area of the triangle, using formula:

$\star\:\boxed{\sf{\purple{\dfrac{1}{2} \bigg[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\bigg]}}}$

Here,

$\frak{Values}\begin{cases}\sf{x_1 = 3}\\\sf{x_2 = 1}\\\sf{x_3 = -2}\\\sf{y_1 = -4}\\\sf{y_2 = 6}\\\sf{y_3 = 3}\end{cases}$

☯ Substituting given values in the formula,

$:\implies\sf Area_{(Triangle)} = \dfrac{1}{2} \bigg[3(6-3) + 1(3+4) + 2(-4 -6) \bigg] \\\\\\:\implies\sf Area_{(Triangle)} = \dfrac{1}{2} \bigg[9 + 7 + 20 \bigg] \\\\\\:\implies\sf Area_{(Triangle)} = 36 \times \dfrac{1}{2} \\\\\\:\implies\boxed{\frak{\pink{ Area_{(Triangle)} = 18 \ square \ units}}}$

$\therefore\:{\underline{\sf{The \: area \: of \: the \: triangle \: is\; {\textsf{\textbf{18 square units}}}.}}}$

Answer 2

Step-by-step explanation:

Diagram:-

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.3,3.2){ \bf (3,-4) }\put(-0.2,-0.3){ \bf (-2,3) }\put(5.2,-0.3){ \bf (1,6) }\end{picture}$

Given:-

The following points are vertices of a triangle

• (3,-4)
• (1,6)
• (-2,3)

To find:-

Area of the triangle

Solution:-

Here

$\sf Values\begin {cases}\star\sf x_1=3,y_1=-4 \\ \star\sf x_2=1,y_2=6 \\ \star\sf x_3=-2,y_3=3\end {cases}$

According to co-ordinate geometry

$\fbox{\sf Area:of\:triangle=\dfrac {1}{2}\left [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right]}$

$:\implies\sf Area_{(Triangle)} = \dfrac{1}{2} \big[3(6-3) + 1(3+4) + 2(-4 -6) \big] \\\\\\\quad\qquad\:\implies\sf Area_{(Triangle)} = \dfrac{1}{2} \big[9 + 7 + 20 \big] \\\\\\\qquad\quad\:\implies\sf Area_{(Triangle)} = 36 \times \dfrac{1}{2} \\\\\\\quad\qquad\:\implies\underline{\boxed{\sf{ Area_{(Triangle)} = 18 \ square \ units}}}$

Formulas of Areas:-

$\\ \star\sf Square=(side)^2 \\ \star\sf Rectangle=Length\times Breadth \\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2$