Question

It is considered a rectangle with equations of 2 sides : `3x-2y-5=0`, `2x+3y+7=0` and a tip A(-2,1). How to determine `S_ABC` ?

Answer 1

`S_(ABC)=3" sq.unit"`.

Explanation:

Method I :

Call the rectangle in question `ABCD,` choosing the vertices in

the anti-clockwise direction.

Observe that,

(1) : The given eqns. represent `bot` **lines.

(2) : The co-ords. of the vertex `A` do not satisfy the given eqns.

In accordance of the above observations, we select,

`CD : 3x-2y-5=0, and, CB : 2x+3y+7=0`.

Since, `{C}=CDnnCB`, solving the resp. eqns., we get

`C=C(1/13,-31/13)`.

Our next target is to find `B`.

Note that, `B in BA |\| CD`.

`BA |\| CD rArr"we may suppose, "BA : 3x-2y+k=0,`

where, `k in RR-{-5}`, is to be determined.

`A=A(-2,1) in BA rArr 3(-2)-2(1)+k=0 rArr k=8`.

`:. BA : 3x-2y+8=0`.

Now, `{B}=CBnnBA,` solving the resp. eqns, we have,

`B=B(-38/13,-5/13)`.

To sum up, we have,

`A(-2,1),B(-38/13,-5/13),C(1/13,-31/13)`.

`:. S_(ABC)=1/2|d|, d=|(-2,1,1),(-38/13,-5/13,1),(1/13,-31/13,1)|`,

`=-2{-5/13-(-31/13)}-1{-38/13-1/13}+1{1178/169-(-5/169)}`,

`=-2(2)-1(-3)+1(7),`

`=-4+3+7,`

`rArr d=6`.

Finally, `S_(ABC)=1/2*6=3" sq.unit"`.

Enjoy Maths., and Spread the Joy!

Answer 2

`3" sq.unit"`.

Explanation:

Method II :

We have, `B=B(-38/13,-5/13)`, as in Method I.

`A=A(-2,1)`. So, using the Distance Formula,

`BA^2={(-2-(-38/13))^2+(1-(-5/13)^2}`,

`=(12/13)^2+(18/13)^2=(6/13)^2(2^2+3^2)`.

`rArr BA=6/sqrt13`.

Knowing that, `DeltaABC` is right- `/_^(ed)`,

`S_(ABC)=1/2*CB*BA`, where,

`CB=DA.........................[because ABCD" is rectangle]"`,

`"="the bot-"distance from point "A" to line "CD`,

`=|3(-2)-2(1)-5|/sqrt(3^2+(-2)^2)`.

`rArr CB=sqrt13`.

`rArr S_(ABC)=1/2*sqrt13*6/sqrt13=3" sq. unit"`, as before!

Enjoy Maths., and Spread the Joy!