Question

The area of a triangle is 3/2 square units.Two of its vertices are the points A(2,-3) and B(3,-2), the centroid of the triangle lies on the line 3x-y-2=0 then third vertex C is​

Answer 1

$\huge{\underline{\sf{\red{Answer\leadsto (-8,-11)}}}}$

Given:

Coordinates Of Triangle $\implies$ A(2,-3),B(3,-2) and C(h,k)

Also We have a line 3x-y-2=0 which passes through the Centroid

Area of Triangle = $\dfrac{3}{2}$

______________________________________

$\huge{\underline{\underline{\green{\tt{Formula}}}}}$

Centroid (x,y) = $\dfrac{x_1+x_2+x_3}{3} ,\dfrac{y_1+y_2+y_3}{3}$

Area of Triangle = $x_1(y_2-y_3) + x_3(y_1-y_2) +x_2(y_3-y_1)$

______________________________________

SoluTion:

$Centroid (x,y) = \dfrac{2+3+h}{3} ,\dfrac{-3-2+k}{3} \\ \\ \dfrac{h+5}{3} ,\dfrac{k-5}{3}$

The Equation given will satisfy centroid coordinates since it is passing through it

$3\bigg(\dfrac{h+5}{\cancel{3}}\bigg) -\dfrac{k-5}{3} -2=0 \\ \\ h+5= \dfrac{k-5+6}{3} \\ \\ 3h+15=k+1 \\ \\ \boxed{3h-k+14=0} ......\large{\boxed{1}}$

Now We know that we Area is 3/2 sq units.

Therefore

$Area = \dfrac{3}{\cancel{2}} =\dfrac{1}{\cancel{2}} \bigg( x_1(y_2-y_3) + x_3(y_1-y_2) +x_2(y_3-y_1)\bigg) \\ \\ \implies \bigg( 2(-2-k) + h(-3+2) + 3(k+3)\bigg) \\ \\ \implies -2k-4 -3h+2h +3k +9 \\ \\ \implies k-h+5 =3 \\ \\ \implies \boxed{h-k=2}.......\large{\boxed{2}}$

Now using Equation $\boxed{1} and \boxed{2}$

$\implies$ 3(k+2) -k +14=0

$\implies$ 3k+6=-14

$\implies$$\huge{\boxed{k=-10}}$

$\implies$ h=k+2

$\implies$ $\huge{\boxed{h=-8}}$

$\leadsto (h,k) = (x,y) = (-8,-11)$

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Answer 2

Answer:

Given:

Coordinates Of Triangle \implies⟹ A(2,-3),B(3,-2) and C(h,k)

Also We have a line 3x-y-2=0 which passes through the Centroid

Area of Triangle = \dfrac{3}{2}

2

3

______________________________________

\huge{\underline{\underline{\green{\tt{Formula}}}}}

Formula

Centroid (x,y) = \dfrac{x_1+x_2+x_3}{3} ,\dfrac{y_1+y_2+y_3}{3}

3

x

1

+x

2

+x

3

,

3

y

1

+y

2

+y

3

Area of Triangle = x_1(y_2-y_3) + x_3(y_1-y_2) +x_2(y_3-y_1)x

1

(y

2

−y

3

)+x

3

(y

1

−y

2

)+x

2

(y

3

−y

1

)

______________________________________

SoluTion:

\begin{lgathered}Centroid (x,y) = \dfrac{2+3+h}{3} ,\dfrac{-3-2+k}{3} \\ \\ \dfrac{h+5}{3} ,\dfrac{k-5}{3}\end{lgathered}

Centroid(x,y)=

3

2+3+h

,

3

−3−2+k

3

h+5

,

3

k−5

The Equation given will satisfy centroid coordinates since it is passing through it

\begin{lgathered}3\bigg(\dfrac{h+5}{\cancel{3}}\bigg) -\dfrac{k-5}{3} -2=0 \\ \\ h+5= \dfrac{k-5+6}{3} \\ \\ 3h+15=k+1 \\ \\ \boxed{3h-k+14=0} ......\large{\boxed{1}}\end{lgathered}

3(

3

h+5

)−

3

k−5

−2=0

h+5=

3

k−5+6

3h+15=k+1

3h−k+14=0

......

1

Now We know that we Area is 3/2 sq units.

Therefore

\begin{lgathered}Area = \dfrac{3}{\cancel{2}} =\dfrac{1}{\cancel{2}} \bigg( x_1(y_2-y_3) + x_3(y_1-y_2) +x_2(y_3-y_1)\bigg) \\ \\ \implies \bigg( 2(-2-k) + h(-3+2) + 3(k+3)\bigg) \\ \\ \implies -2k-4 -3h+2h +3k +9 \\ \\ \implies k-h+5 =3 \\ \\ \implies \boxed{h-k=2}.......\large{\boxed{2}}\end{lgathered}

Area=

2

3

=

2

1

(x

1

(y

2

−y

3

)+x

3

(y

1

−y

2

)+x

2

(y

3

−y

1

))

⟹(2(−2−k)+h(−3+2)+3(k+3))

⟹−2k−4−3h+2h+3k+9

⟹k−h+5=3

⟹

h−k=2

.......

2

Now using Equation \boxed{1} and \boxed{2}

1

and

2

\implies⟹ 3(k+2) -k +14=0

\implies⟹ 3k+6=-14

\implies⟹ \huge{\boxed{k=-10}}

k=−10

\implies⟹ h=k+2

\implies⟹ \huge{\boxed{h=-8}}

h=−8

\leadsto (h,k) = (x,y) = (-8,-11)⇝(h,k)=(x,y)=(−8,−11)

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