Question

find the area of triangle formed by the points A(1,2) ,B(2,10) and C(3,10) which are collinear​

Answer 1

Correct Question

⇒Find the area of triangle formed by the points A(1,2) ,B(2,10) and C(3,10)

     

Solution

Given Point

⇒A(1,2) ,B(2,10) and C(3,10)

Formula

⇒ΔABC = 1/2{x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)}

Now we have

⇒x₁=1 ,y₁=2 ,x₂ = 2 ,y₂ = 10 , x₃=3 and y₃ = 10

Put the value on formula

⇒ar(ΔABC) = |1/2{1(10- 10) + 2(10 - 2) + 3(2 - 10)}|

⇒ar(ΔABC) = |1/2{1(0) + 2(8) + 3(-8)}|

⇒ar(ΔABC)  = |1/2{0+16 - 24}|

⇒ar(ΔABC) = |1/2{-8}|

⇒ar(ΔABC) = |-4|

⇒ar(ΔABC) = 4 sq unit

Answer  

⇒ar(ΔABC) = 4 sq unit

Answer 2

${\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}$

Given that: The points are given that form a triangle are A(1,2) ,B(2,10) and C(3,10) which are collinear.

To find: The area of triangle formed by the points A(1,2) ,B(2,10) and C(3,10) which are collinear

Solution: The area of triangle formed by the points A(1,2) ,B(2,10) and C(3,10) which are collinear = 4 unit sq.

Using concept: Formula to find out area of triangle according to the question.

Using formula:

${\small{\underline{\boxed{\sf{\dfrac{1}{2} [x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)]}}}}}$

Where,

• ${\sf{x_1 \: is \: 1}}$
• ${\sf{y_2 \: is \: 10}}$
• ${\sf{y_3 \: is \: 10}}$
• ${\sf{x_2 \: is \: 2}}$
• ${\sf{y_1 \: is \: 2}}$
• ${\sf{x_3 \: is \: 3}}$

Full solution:

${\small{\underline{\boxed{\sf{\dfrac{1}{2} [x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)]}}}}} \\ \\ :\implies \sf \dfrac{1}{2} [x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)] \\ \\ :\implies \sf \dfrac{1}{2}[1(10-10) + 2(10-2) + 3(2-10) \\ \\ :\implies \sf \dfrac{1}{2}[1(0) + 2(8) + 3(-8) \\ \\ :\implies \sf \dfrac{1}{2}(0) + (16) (-24) \\ \\ :\implies \sf \dfrac{1}{2} (0 + 16 - 24) \\ \\ :\implies \sf \dfrac{1}{2}(16-24) \\ \\ :\implies \sf \dfrac{1}{2}(-8) \\ \\ :\implies \sf -4 \: unit \: sq. \\ \\ :\implies \sf Area \: = 4 \: unit \: sq.$

Additional information:

$\underline{\bigstar\:\textsf{Distance Formula\; :}}$

Distance formula is used to find the distance between two given points.

${\underline{\boxed{\sf{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$ ⠀

$\underline{\bigstar\:\textsf{Section Formula\; :}}$

Section Formula is used to find the co ordinates of the point(Q) which divides the line segment joining the points (B) and (C) internally or externally.

${\underline{\boxed{\sf{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$ ⠀

$\underline{\bigstar\:\textsf{Mid Point Formula\; :}}$

Mid Point formula is used to find the mid points on any line.

${\underline{\boxed{\sf{\quad \Bigg(\dfrac{x_1 + x_2}{2} \; or\; \dfrac{y_1 + y_2}{2} \Bigg)\quad}}}}$