Question

Find the area of the triangle whose vertices are A(10, 6), B(2, 5) and C(-1, 3) units (b) 24.5 sq. sq. units (c) 7 sq. units units sq. (a) 12.5 (d) 6.5​

Answer 1

Given: The vertices of the triangle are A(10, 6), B(2, 5) and C(–1, 3).

⠀☢ To Calculate the area of the triangle with given three vertices the formula is Given by —

$\star\;\underline{\boxed{\frak{Area_{\:(triangle)} = \dfrac{1}{2}\Bigg\{x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\Bigg\}}}}$

$\sf{Here}\begin{cases}\sf{\quad x_1, \; y_1 = \bf{10, 6}}\\\\\sf{\quad x_2,\; y_2 = \bf{2, 5}}\\\\\sf{\quad x_3, \;y_3 = \bf{-1, 3}}\end{cases}$

⠀

⠀⠀$\underline{\bf{\dag} \:\mathfrak{Substituting \:values~in~the~given~formula~: :}}$⠀⠀⠀⠀

$:\implies\sf\dfrac{1}{2}\Bigg\{10(5 - 3) + 2(3 - 6) +(-1)(6 - 5)\Bigg\} \\\\\\:\implies\sf \dfrac{1}{2}\Bigg\{10 \times 2 + 2 \times (-3) - 1 \times 1\Bigg\}\\\\\\:\implies\sf \dfrac{1}{2}\Bigg\{ 20 - 6 - 1\Bigg\}\\\\\\:\implies\sf\dfrac{1}{2}\Bigg\{20 - 7\Bigg\} \\\\\\:\implies\sf \dfrac{1}{2} \times 13\\\\\\:\implies\underline{\boxed{\pmb{\frak{\red{6.5}}}}}\;\bigstar$

$\therefore\:\underline{\textsf{Hence, Area of the triangle is \textbf{Option d) 6.5 sq. units}}}.$

Answer 2

Given:-

Coordinate A=(10,-6)

Coordinate B=(2,5)

Coordinate C=(-1,3)

To Find:-

According to question we have to find the area of triangle.

Solution:-

(10,-6;(2,5);(-1,3)

_________

By using:-

x1 y1, x2 y2, x3 y3

Area of triangle=$\dfrac{1}{2}$[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

∴Area of triangle=$\dfrac{1}{2}$[10(5-3)+2(3+6)-1(-6-5)]

∴Area of triangle=$\dfrac{1}{2}$[10×(2)+2(9)-1×11]

∴Area of triangle=$\dfrac{1}{2}$[20+18+11]

∴ Area of triangle=$\dfrac{1}{2}$[49]

∴ Area of triangle=24.5 sq units

______________

Hence$<strong> </strong>\sf \pink{Option \ [B]24.5}$is the correct answer✔️