Question

If A(6.-1), B(1.3) and C(k, 8) are three points such that AB=BC, find the value of k.
find the value​

Answer 1

Step-by-step explanation:

Given:-

A(6.-1), B(1.3) and C(k, 8) are three points such that AB=BC

To find:-

Find the value of k ?

Solution:-

Given points are A(6.-1), B(1.3) and C(k, 8)

Given that AB = BC

I) Length of AB:-

Let (x1, y1)=A(6,- 1)=>x1 = 6 and y1 = -1

Let (x2, y2)=B(1,3)=>x2=1 and y2=3

We know that the distance between two points

( x1, y1) and (x2,y2)is√[(x2-x1)^2+(y2-y1)^2] units

=>AB = √[(1-6)^2+(3-(-1))^2]

=>AB= √[(-5)^2+(3+1)^2]

=>AB = √[25+(4)^2]

=>AB=√(25+16)

AB=√41 units --------(1)

ii)Length of BC:-

Let (x1, y1)=B(1,3)=>x1=1 and y1=3

Let (x2, y2)=C(k,8)=>x2=k and y2=8

We know that the distance between two points

( x1, y1) and (x2,y2)is√[(x2-x1)^2+(y2-y1)^2] units

BC =√[(k-1)^2+(8-3)^2]

=>BC=√[(k-1)^2+5^2]

=>BC= √[k^2-2k+1+25]

BC=√(k^2-2k+26) -----------(2)

From (1)&(2)

=>(1)=(2)

=>AB=BC

=>√41 = √(k^2-2k+26)

On squaring both sides then

(√41 )^2= [√(k^2-2k+26)]^2

=>41 = k^2-2k+26

=>k^2-2k+26-41 = 0

=>k^2-2k-15 = 0

=>k^2+3k-5k-15 = 0

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

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

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

=>k = -3 or k=5

Answer:-

The values of k for the given problem are -3 and 5

Used formula:-

the distance between two points

( x1, y1) and (x2,y2)is√[(x2-x1)^2+(y2-y1)^2] units