Question

verify the formula of area of triangle (in co-ordinate geometry) with the help of the formula of plane geometry.​

Answer 1

Answer:

1/2×b×h shdirvdjd

Step-by-step explanation:

good

Answer 2

Answer:

Step-by-step explanation:

Let us assume a triangle PQR, whose coordinates P, Q, and R are given as (x1, y1), (x2, y2), (x3, y3), respectively.

From the figure, the area of a triangle PQR, lines such as  

←→   ←→        ←→

Q

A

,  P

B  and  R

C  are drawn from Q, P, and R, respectively perpendicular to the x-axis.

Now, three different trapeziums are formed such as PQAB, PBCR, and QACR in the coordinate plane.

Now, calculate the area of all the trapeziums.

Therefore, the area of ∆PQR is calculated as Area of ∆PQR=[Area of trapezium PQAB + Area of trapezium PBCR] -[Area of trapezium QACR] —(1)

Finding Area of a Trapezium PQAB

We know that the formula to find the area of a trapezium is

Since Area of a trapezium = (1/2) (sum of the parallel sides)×(distance between them)

Area of trapezium PQAB = (1/2)(QA + PB) × AB

QA =

PB =

AB = OB – OA = –

Area of trapezium PQAB = (1/2)( $y_{1}$+ $y_{2}$)($x_{1}$ – $x_{2}$) —-(2)

Finding Area of a Trapezium PBCR

Area of trapezium PBCR =(1/2) (PB + CR) × BC

PB =  

CR =

BC = OC – OB = –

Area of trapezium PBCR =(1/2) (+ )(– ) —-(3)

Finding Area of a Trapezium QACR

Area of trapezium QACR = (1/2) (QA + CR) × AC

QA =

CR =

AC = OC – OA = –

Area of trapezium QACR =(1/2)( + ) ( – )—-(4)

Substituting (2), (3), and (4) in (1),

Area of ∆PQR = (1/2)[( + )( – ) + (  + ) – ) – ( + ) ( – )]

                         = (1/2) [$x_{1}$ ($y_{2}$ – $y_{3}$ ) + $x_{2}$ ($y_{3}$ – $y_{1}$  ) + $x_{3}$(