Question

find the area of triangle whose vertics are 2,-4 and -1, 0 2,3​

Answer 1

Answer:

13/2

Step-by-step explanation:

Area of ∆=1/2 | {(2×0)-(-1×-4)} + {(-1×3)-(2×0)} + {(2×-4)-(2×3)} |

=1/2 |(0+4) + (-3-0) + (-8-6)|

=1/2 |4-3-14|

=1/2 |-13|

Since, area can't be in negative

Therefore, Area of ∆= 13/2

Answer 2

Answer :-

21/2 sq.units

Given to find the area of rectangle whose vertices are :-

• ( 2, -4)
• (-1 , 0)
• (2 , 3)

Diagram :-

$\setlength{\unitlength}{2.5mm}\begin{picture}(0,0)\linethickness{0.3mm}\qbezier(0,0)(0,0)(8,17)\qbezier(0,0)(0,0)(18,0)\qbezier(18,0)(18,0)(8,17)\put(8,17.8){\sf A}\put( - 1, - 1){\sf C}\put(18,-1){\sf B}\put(8, - 1.5){\sf (-1,0)}\put(15,8.1){\sf (-2,3)}\put( - 0,8.1){\sf (2, -4)}\end{picture}$

Solution :-

By using area of triangle formula We can find the area of triangle

Formulae  Implemented :-

$\dfrac{1}{2} \bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|$

$x_1 = 2 \\x_2= -1 \\x_3 = 2$

$y_1 = -4\\y_2 = 0\\y_3= 3$

Substituting the values ,

$\dfrac{1}{2} \bigg|2(0-3)-1(3+4) +2(-4-0)\bigg|$

$\dfrac{1}{2} \bigg|2(-3)-1(7) +2(-4)\bigg|$

$\dfrac{1}{2} \bigg|-6-7 -8\bigg|$

$\dfrac{1}{2} \bigg|-13-8\bigg|$

$\dfrac{1}{2} \bigg|-21\bigg|$

$\dfrac{1}{2} (21)$

$\dfrac{21}{2} sq.units$

Method -2 :-

By using determinants we can solve The required formula is

$\dfrac{1}{2} \left|\begin{array}{cc}x_1&y_1&\\x_2&y_2&\end{array}\right|+\left|\begin{array}{cc}x_2&y_2&\\x_3&y_3&\end{array}\right|+\left|\begin{array}{cc}x_3&y_3&\\x_1&y_1&\end{array}\right|$

Substituting the values ,

$\dfrac{1}{2} \left|\begin{array}{cc}2&-4&\\-1&0&\end{array}\right|+\left|\begin{array}{cc}-1&0&\\2&3&\end{array}\right|+\left|\begin{array}{cc}2&3&\\2&-4&\end{array}\right|$

By applying ad-bc formula

$\dfrac{1}{2} \bigg|2(0)-(-4)(-1)\bigg|+\bigg|(-1)(3)-(2)(0)\bigg|+\bigg|2(-4)-2(3)\bigg|$

$\dfrac{1}{2} \bigg|-4\bigg|+\bigg|-3\bigg|+\bigg|-8-6\bigg|$

$\dfrac{1}{2} \bigg|-4\bigg|+\bigg|-3\bigg|+\bigg|-14\bigg|$

$\dfrac{1}{2} (4+3+14)$

$\dfrac{21}{2} sq.units$

So, area of triangle formed by those points were 21/2 sq.units