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 About the author Harper
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] ⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ Reply
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 Reply
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]
⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
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