Question

if a(6 -1) B,(1,3), C(k,8) are three points such that AB=BC find the value of k

Answer 1

Given :-

• A(6,-1)

• B(1,3)

• C(k,8)

• AB = BC

To Find :-

• Value of k

Answer :-

• -3 , 5

Solution :-

$\bf{ \underline{ \boxed{ \red{Distance\:Formula}}}}$

$\huge\sf{1}$

$\sf{Distance\:=\:\sqrt{\ (\ x_{2}-\ x_{1})^{2}+\ (\ y_{2}-\ y_{1})^{2}}}$

$\sf{AB\:=\:\sqrt{\ (1-6)^{2}+\ (3-(-1))^{2}}}$

$\sf{AB\:=\:\sqrt{\ (-5)^{2}+\ (4)^{2}}}$

$\sf{AB\:=\:\sqrt{25+16}}$

$\sf{AB\:=\:\sqrt{41}}$

$\huge\sf{2}$

$\sf{Distance\:=\:\sqrt{\ (\ x_{2}-\ x_{1})^{2}+\ (\ y_{2}-\ y_{1})^{2}}}$

$\sf{BC\:=\:\sqrt{\ (k-1)^{2}+\ (8-3)^{2}}}$

$\sf{BC\:=\:\sqrt{\ k^{2}+1-2k+\ (5)^{2}}}$

$\sf{BC\:=\:\sqrt{\ k^{2}+1-2k+25}}$

$\sf{BC\:=\:\sqrt{\ k^{2}-2k+26}}$

$\bf{A.T.Q}$

$\sf{AB\:=\:BC}$

$\bf{Put\:value\:of\:AB\:and\:BC}$

$\sf{\sqrt{41}\:=\:\sqrt{\ k^{2}-2k+26}}$

$\bf{Root\:will \:cancel\:from\:each\:side}$

$\sf{\ k^{2}-2k+26-41\:=\:0}$

$\sf{\ k^{2}-2k-15\:=\:0}$

$\sf{\ k^{2}+3k-5k-15\:=\:0}$

$\sf{k(k+3)-5(k+3)\:=\:0}$

$\sf{(k+3)(k-5)\:=\:0}$

$\sf{k\:=\:-3\:,\:5}$

Answer 2

$\huge \red{ \boxed{question}}$

if a(6,-1), B(1,3) ,C(k,8) are three points such that AB=Bc find the value of K

$\huge \blue{ \underline{answer}}$

The given points are A(6, −1), B(1, 3) and C(k, 8). Also, it is given that AB = BC. Therefore, the value of k will be either 5 or −3.

$\huge \red \bigstar \green{explanation} \red \bigstar$

AB=BC²

=[(6-1)10] + (-1-3)²] =[(k-1)² +(8-3)²]

= (25+16) = (ki+1-2k+25)

=k²-2k-15=0

=k²-5k+3k-15=0

=k(k-5) +3(k-5) =0

=(k-5) (k+3) =0

=(k-5)=0 OR (k+3) =0

=>K=5 OR K=-3

I HOPE ITS HELP U ❤️