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Questions

August 2021 1 22 Report

[tex]\tt{QUESTION}[/tex]

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

[tex]\shortrightarrow[/tex]
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Answers & Comments

BRAINLYADDICTOR

★FIND:

The value of 'k'

★GIVEN,

In Isosceles (triangle) ⛛ABC three vertices are

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

★SOLUTION:

In isosceles triangle(⛛) two sides are equal.

So,

AB=AC

$= > \sqrt{(2 - ( -4 )) {}^{2} +( - 1 - 1) {}^{2} } = \sqrt{(1 - ( - 4)) {}^{2} + (k - 1) {}^{2} } \\ cancelling \: square \: root \: on \: b.s \\ = > (2 + 4) {}^{2} + ( - 2) {}^{2} = (1 + 4) {}^{2} + (k - 1) {}^{2} \\ = > 6 {}^{2} + 4 = 5 {}^{2} + k {}^{2} + 1 - 2k \\ = > 36 + 4 = 25 + 1 + k {}^{2} - 2k \\ = > 40 = 26 + k {}^{2} - 2k \\ = > k {}^{2} - 2k + 26 - 40 = 0 \\ = > k {}^{2} - 2k - 14 = 0 \\ here \\ a = 1.b = - 2.c = - 14 \\ k = - b + - \sqrt{b {}^{2} - 4ac} /2a \\ = > - ( - 2) + - \sqrt{( - 2) {}^{2} - 4(1)( - 14) } /2(1) \\ = > 2 + - \sqrt{4 + 56} /2 \\ = > 2 + - \sqrt{60}/2 \\ = > 2 + - \sqrt{15 \times 4} /2 \\ = > 2 + - 2 \sqrt{15} /2 \\ = > 1 + - \sqrt{15} \\ k = 1 + \sqrt{15} (or)k = 1 - \sqrt{15}$