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 [tex]\triangle [/tex] ABC :

    [tex]\qquad \longmapsto \sf AB = AC \\ [/tex]

    ━━━━ By Using the Distance Formula :

    [tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex]

    [tex]\qquad \dag\:\:\bigg\lgroup \sf{ Distance \:: \sqrt {(x_2 – x_1 )^2+ (y_2 – y_1 )^2} }\bigg\rgroup \\\\[/tex]

    ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex]

    [tex]\qquad \longmapsto \sf \sqrt { (1 – 4)^2 + (-1 -3 )^2} =\sqrt { (7-4)^2 + (k-3)^2 } \\\\ [/tex]

    [tex]\qquad \longmapsto \sf \sqrt { ( – 3)^2 + ( -4 )^2} =\sqrt { (3)^2 + (k-3)^2 } \\\\ [/tex]

    [tex]\qquad \longmapsto \sf \sqrt { 9 + 16 } =\sqrt { (3)^2 + (k-3)^2 } \\\\ [/tex]

    [tex]\qquad \longmapsto \sf \sqrt { 9 + 16 } =\sqrt { 9 + (k-3)^2 } \\\\ [/tex]

    By Squaring Both side we get ,

    [tex]\qquad \longmapsto \sf 9 + 16 = 9 + (k-3)^2 \\\\ [/tex]

    [tex]\qquad \longmapsto \sf 25 = 9 + (k-3)^2 \\\\ [/tex]

    [tex]\qquad \longmapsto \sf 25 – 9 = (k-3)^2 \\\\ [/tex]

    [tex]\qquad \longmapsto \sf 16 = (k-3)^2 \\\\ [/tex]

    As , We know that ,

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

    ━By Using this Algebraic indentity :

    [tex]\qquad \longmapsto \sf 16 = (k-3)^2 \\\\ [/tex]

    [tex]\qquad \longmapsto \sf 16 = k^2 + 9 – 6k \\\\ [/tex]

    [tex]\qquad \longmapsto \sf 16 = k^2 + 3^2 – 2\times 3 \times k \\\\ [/tex]

    [tex]\qquad \longmapsto \sf 16 = k^2 + 3^2 – 6 \times k \\\\ [/tex]

    [tex]\qquad \longmapsto \sf 16 = k^2 + 3^2 – 6k \\\\ [/tex]

    [tex]\qquad \longmapsto \sf 16 = k^2 + 9 – 6k \\\\ [/tex]

    [tex]\qquad \longmapsto \sf 0 = k^2- 6k + 9 – 16 \\\\ [/tex]

    [tex]\qquad \longmapsto \sf 0 = k^2- 6k – 7 \\\\ [/tex]

    ━By Using Sum – Product Pattern :

    [tex]\qquad \longmapsto \sf 0 = k^2- 7k + k – 7 \\\\ [/tex]

    ━By Finding Common term :

    [tex]\qquad \longmapsto \sf 0 = k(k-7)\: + 1(k- 7) \\\\ [/tex]

    ━Now By Rewrite in Factored Term :

    [tex]\qquad \longmapsto \sf 0 = ( k – 7 ) ( k + 1) \\\\ [/tex]

    [tex]\qquad \longmapsto \sf k = 7 \:or\:\:-1 \\\\ [/tex]

    ━[ Distance cannot be calculated in ” – ” ]

    [tex]\qquad \longmapsto \frak{\underline{\purple{\:k = 7 }} }\bigstar \\[/tex]

    Therefore,

    ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm {\:The\:Value \:of\:k \:is\:\bf{7}}}}\\[/tex]

    ⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

  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

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