Question

Three vertices of a triangle are A=(4,3) B=(1,-1) C=(7,k) The value of k such that the centroid, orthocentre, circumcentre and incentre lie on same line

HINT :- answer should be 7​

Answer 1

EXPLANATION.

Three vertices of a triangle,

⇒ A = (4,3)  and  B = (1,-1)  and  C = (7,k).

Centroid, orthocentre, Circumcentre and incentre lie on same line.

As we know that,

Note :

In an isosceles triangle Centroid, orthocentre, Circumcentre and incentre lie on same line.

As we know that,

Formula of :

Distance formula :

⇒ √(x₂ - x₁)² + (y₂ - y₁)².

⇒ A = (4,3)  & B = (1,-1)  &  C = (7,k).

⇒ AB = AC.

Using distance formula in equation, we get.

⇒ √(1 - 4)² + (-1 - 3)² = √(7 - 4)² + (k - 3)².

⇒ √(-3)² + (-4)² = √(3)² + (k - 3)².

⇒ √9 + 16 = √9 + (k - 3)².

Squaring on both sides, we get.

⇒ 25 = 9 + (k - 3)².

⇒ 25 - 9 = (k - 3)².

⇒ 16 = (k - 3)².

As we know that,

Formula of :

⇒ (x - y)² = x² + y² - 2xy.

Using this formula in equation, we get.

⇒ 16 = k² + 9 - 6k.

⇒ k² - 6k + 9 - 16 = 0.

⇒ k² - 6k - 7 = 0.

Factorizes the equation into middle term splits, we get.

⇒ k² - 7k + k - 7 = 0.

⇒ k(k - 7) + 1 (k - 7) = 0.

⇒ (k - 7)(k + 1) = 0.

⇒ k = 7 and k = - 1.

k = 7 satisfied the equation.

Value of k = 7.

                                                                                                                                             

MORE INFORMATION.

Positions of four points.

Four given points A, B, C, D are vertices of a,

(1) = Square if AB = BC = CD = DA and AC = BD.

(2) = Rhombus if AB = BC = CD = DA and AC ≠ BD.

(3) = Parallelogram if AB = DC ; BC = AD ; AC ≠ BD.

(4) = Rectangle if AB = CD ; BC = DA ; AC = BD.

Note :

(1) = Diagonal of square, rhombus, rectangle and parallelogram always bisect each other.

(2) = Diagonal of rhombus and square bisect each other at right angle.

Answer 2

Given : Three vertices of a triangle are A=(4,3) B=(1,-1) C=(7,k) . The centroid, orthocentre, circumcentre and incentre lie on same straight line .

Exigency to find : The Value of k .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Given that ,

• The centroid, orthocentre, circumcentre and incentre lie on same straight line .

As , We know that ,

━━━━ When centroid , orthocentre , circumcentre & incentre lie on same straight line the the triangle is an Isosceles Triangle .

Therefore ,

⠀⠀⠀We can say that ,

• In $\triangle$ ABC :

$\qquad \longmapsto \sf AB = AC$

━━━━ By Using the Distance Formula :

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ Distance \:: \sqrt {(x_2 - x_1 )^2+ (y_2 - y_1 )^2} }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \longmapsto \sf \sqrt { (1 - 4)^2 + (-1 -3 )^2} =\sqrt { (7-4)^2 + (k-3)^2 }$

$\qquad \longmapsto \sf \sqrt { ( - 3)^2 + ( -4 )^2} =\sqrt { (3)^2 + (k-3)^2 }$

$\qquad \longmapsto \sf \sqrt { 9 + 16 } =\sqrt { (3)^2 + (k-3)^2 }$

$\qquad \longmapsto \sf \sqrt { 9 + 16 } =\sqrt { 9 + (k-3)^2 }$

• By Squaring Both side we get ,

$\qquad \longmapsto \sf 9 + 16 = 9 + (k-3)^2$

$\qquad \longmapsto \sf 25 = 9 + (k-3)^2$

$\qquad \longmapsto \sf 25 - 9 = (k-3)^2$

$\qquad \longmapsto \sf 16 = (k-3)^2$

As , We know that ,

• Algebraic Indentity = ( a - b) ² = a² + b² - 2ab

━By Using this Algebraic indentity :

$\qquad \longmapsto \sf 16 = (k-3)^2$

$\qquad \longmapsto \sf 16 = k^2 + 9 - 6k$

$\qquad \longmapsto \sf 16 = k^2 + 3^2 - 2\times 3 \times k$

$\qquad \longmapsto \sf 16 = k^2 + 3^2 - 6 \times k$

$\qquad \longmapsto \sf 16 = k^2 + 3^2 - 6k$

$\qquad \longmapsto \sf 16 = k^2 + 9 - 6k$

$\qquad \longmapsto \sf 0 = k^2- 6k + 9 - 16$

$\qquad \longmapsto \sf 0 = k^2- 6k - 7$

━By Using Sum - Product Pattern :

$\qquad \longmapsto \sf 0 = k^2- 7k + k - 7$

━By Finding Common term :

$\qquad \longmapsto \sf 0 = k(k-7)\: + 1(k- 7)$

━Now By Rewrite in Factored Term :

$\qquad \longmapsto \sf 0 = ( k - 7 ) ( k + 1)$

$\qquad \longmapsto \sf k = 7 \:or\:\:-1$

━[ Distance cannot be calculated in " - " ]

$\qquad \longmapsto \frak{\underline{\purple{\:k = 7 }} }\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The\:Value \:of\:k \:is\:\bf{7}}}}$

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