Question

ABC is a triangle and G (1,2) is its centroid. If A(2,5) and B(6, b) and C (a,3) , (3) then find a and b. Also find the length of side BC.

Answer 1

Answer:

Explanation:

Given :

• ΔABC is a triangle and G (1,2) is centroid.
• A(2,5) and B(6, b) and C (a,3).

To Find :

• Find the value of a and b.
• Find the length of side BC.

Solution :

★Find the value of a and b.

Applying centroid formula,

$\tt \red{x = \dfrac{x_1 +x_2 + x_3 }{3} \: \: \: \: \: \: \: and \: \: \: \: \: \: \: y = \dfrac{y_1 +y_2 + y_3 }{3} } \\ \\ \\ : \implies \tt\: 1 = \dfrac{2 + 6 + a}{3} \: \: \: \: \: \: \: and \: \: \: \: \: \: 2 = \dfrac{5 + b+ 3}{3} \\ \\ \\ : \implies \tt \: 1 = \dfrac{8 + a}{3} \: \: \: \: \: \: \: and \: \: \: \: \: \: 2 = \dfrac{8 + b}{3} \\ \\ \\ : \implies \tt 3 = 8 + b\: \: \: \: \: \: \: and \: \: \: \: \: \: 6 = 8 + b \\ \\ \\ \implies \tt \purple{a = 5} \: \: \: \: \: \: \: and \: \: \: \: \: \: \purple{b = 2}$

★Find the length of side BC.

Applying distance formula,

$\tt \red{d = \sqrt{(x_2 - x_1) {}^{2} +(y_2 - y_1) {}^{2} } } \\ \\ \\ : \implies \tt \: BC = \sqrt{(5 - 6) {}^{2} + (3 - 2) {}^{2} } \\ \\ \\ : \implies \tt \: BC = \sqrt{( - 1) {}^{2} + (1) {}^{2} } \\ \\ \\ : \implies \tt \: BC = \sqrt{(1 + 1) {}^{2} } \\ \\ \\ : \implies \tt \: BC = \sqrt{(2) {}^{2} } \\ \\ \\ : \implies \tt \purple{BC = 2\: \: units }$

Answer 2

$\large \mathfrak{ \underline{ \color{purple}{Given : }}}$

• $\sf{ \green{ABC \: is \: a \: triangle \: and \: G (1,2) is \: centroid.}}$
• $\sf{ \green{A(2,5) and \: B(6, b) and \: C (a,3).}}$

$\large \mathfrak{ \underline{ \color{navy}{To Find :}}}$

• $\sf{ \orange{The \: value \: of \: a \: and \: b}}$
• $\sf{ \orange{The \: length \: of \: side \: BC.}}$

$\large \mathfrak{ \underline{ \color{green}{Solution : }}}$

➠ Finding the value of a and b.

$\sf{ \color{teal}{Applying centroid formula,}}$

$\begin{gathered} \boxed{\rm \red{x = \dfrac{x_1 +x_2 + x_3 }{3} \: \: \: \:and \: \: \: \: y = \dfrac{y_1 +y_2 + y_3 }{3} }} \\ \\ \\ { \blue{\implies \sf\: 1 = \dfrac{2 + 6 + a}{3} \: \: \: \: \: \: \: and \: \: \: \: \: \: 2 = \dfrac{5 + b+ 3}{3}}} \\ \\ \\ { \blue{\implies \sf\: 1 = \dfrac{8 + a}{3} \: \: \: \: \: \: \: and \: \: \: \: \: \: 2 = \dfrac{8 + b}{3}}} \\ \\ \\ { \blue{ \implies \sf 3 = 8 + b\: \: \: \: \: \: \: and \: \: \: \: \: \: 6 = 8 + b}} \\ \\ \\ { \pink{\implies \sf{a = 5}\: \: \: \: \: \: \: and \: \: \: \: \: \: \ \: b = 2}}\end{gathered}$

➠ Finding the length of side BC.

$\sf{ \color{teal}{Applying distance formula,}}$

$\begin{gathered} \boxed{\rm\red{d = \sqrt{(x_2 - x_1) {}^{2} +(y_2 - y_1) {}^{2} } }} \\ \\ \\{ \blue{ \implies \sf\: BC = \sqrt{(5 - 6) {}^{2} + (3 - 2) {}^{2} }}} \\ \\ \\ { \blue{ \implies \sf \: BC = \sqrt{( - 1) {}^{2} + (1) {}^{2} }}} \\ \\ \\ { \blue{\implies \sf \: BC = \sqrt{(1 + 1) {}^{2}}} } \\ \\ \\ { \blue{\implies \sf\: BC = \sqrt{(2) {}^{2}}} } \\ \\ \\ { \pink{ \implies \sf {BC = 2\: \: units }}}\end{gathered}$