# 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 inc

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​

### 2 thoughts on “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 inc”

1. 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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2. Step-by-step explanation:

centroid, arthocenter, incenter &

circumcenter

of AABC lie on same line so AABC is

isosceles A.

. AB = AC

AB2 = AC2

V(1– 4)? + (–1 – 3)? =

(7 — 4)2 + (k – 3)2

= 9+ 16 = 9+ (k – 3)?

= 16(k 3)²

..k – 3 = 4or k – 3 = -4

+ k = 7 k = -1