Question

$\tt{QUESTION}$

triangle ABC has vertices A (-4,1) , B (2,-1),C (1,K) then no of possible values of K is so that triangle ABC is isosceles
(a) 2
(b) 1
(c) 5
(d) 6

$\shortrightarrow$
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Answer 1

● Given :

ABC is a isosceles triangle with vertices -

A ( - 4 , 1 )

A ( - 4 , 1 ) B ( 2 , - 1 )

A ( - 4 , 1 ) B ( 2 , - 1 ) C ( 1 , k )

● To find :

Number of all possible values of k .

● Answer :

option ( c ) $\longrightarrow$ 5

● Solution:

Triangle ABC is isosceles , it means any of the two sides are equal.

there may be 3 cases .

Case 1 : AB = BC

Case 2 : BC = AC

Case 3 : AC = AB

Case 1 :-

when AB = BC

$\sqrt{ {(2 + 4)}^{2} + {( - 2)}^{2} } = \sqrt{ {( - 1)}^{2} + {(k + 1)}^{2} }$

$36 + 4 = 1 + {k}^{2} + 1 + 2k$

$= {k}^{2} + 2k - 38$

( By using quadratic formula )

$k = \frac{ - 2 + - \sqrt{4 - 4 \times 38 \times 1} }{2}$

$= \frac{ - 2 + - \sqrt{4 + 152} }{2}$

$= \frac{ - 2 + - \sqrt{15} }{2}$

Here , in case 1 , two values of k are possible , that are -

$\frac{ - 2 + \sqrt{15} }{2}$

and

$\frac{ - 2 - \sqrt{15} }{2}$

Case 2 :-

when BC = AC

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

$1 + {k}^{2} + 1 + 2k = 25 + {k}^{2} + 1 - 2k$

$4k = 24$

$k = 6$

Here, in case 2 , only one value of k is possible that is 6 .

Case 3 :-

when AC = AB

$\sqrt{ {(5)}^{2} + {(k - 1)}^{2} } = \sqrt{ {(6)}^{2} + {( - 2)}^{2} }$

$25 + {k}^{2} + 1 - 2k = 36 + 4$

${k}^{2} - 2k - 14 = 0$

after solving ,

$k = \frac{2 + - \sqrt{60} }{2}$

$k = 1 + - \sqrt{15}$

Here , in case 3 , two values of k are possible that are -

$1 + \sqrt{15}$

and

$1 - \sqrt{15}$

Total number of possible values = number of possible values in Case 1 + number of possible values in case 2 + number of possible values in case 3

=> 2 + 1 + 2 = 5

Hence , option ( c ) is correct .

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Hope it will definitely help you .

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Answer 2

Answer:

There are 5 possible value for k .

Step-by-step explanation:

Explanation :

Given , ABC is a triangle which has vertices,

A (-4,1) , B(2,-1) and (1,k).

So, if a triangle is an isosceles than its any two side be equal to each other .

Such as , AB = BC or BC = AC or AC = AB

So , there are three case .

Therefore , by using distance formula we first find AB , BC and AC

Step1:

AB = $\sqrt{(2+4)^{2} +(-1-1)^{2} } = \sqrt{36+4} = \sqrt{40}$

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

AC = $\sqrt{(1+5)^{2} +(k-1)^{2} } = \sqrt{k^{2} -2k+24}$

Now , in case 1 which is AB = BC

∴$\sqrt{40} = \sqrt{k^{2} +2k+2}$  

Squaring both side  we get ,

$40 = k^{2} +2k +2$

⇒38 = $k^{2} +2k$ ⇒$k^{2} +2k +38 = 0$

k = $\frac{-2+-\sqrt{4+4.38.1} }{2} = \frac{-2+\sqrt{15} }{2} and \frac{-2-\sqrt{15} }{2}$

There are two value of k

Step2:

Case 2 : BC = AC

$\sqrt{k^{2}+2k+2 }$ = $\sqrt{k^{2} -2k+24}$

Squaring  both side we get ,

$k^{2}+2k+2 = k^{2} -2k+24$

⇒4k = 24 ⇒k = 6

Only , one value of k is possible ,

Step3:

Case 3: AC = AB

$\sqrt{5^{2} +(k+1)} = \sqrt{40}$

Squaring both side

$25 +k^{2} +1-2k = 40$

⇒ $k^{2} - 2k -14 =0$

Solving this equation ,

k = 1+$\sqrt{15}$ , 1-$\sqrt{15}$

In this case there are two values of k

So ,total number of possible values of k

case (i) +case (ii)+case (iii) = 2+1+2 = 5

Final answer :

Hence ,there are  5 possible value for k and the correct answer is option (c) .