Question

if the system of linear equations 3x+2y-4=0 and kx-y-3= 0 represents intersecting lines then find k ?​

Answer 1

Answer:

k=4

Step-by-step explanation:

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Answer 2

$\huge\boxed{\underline{\underline{\green{Solution}}}}$

FORMULA TO BE IMPLEMENTED :

A pair of Straight Lines

$\displaystyle \: a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0$

is said to represents pair of intersecting lines if

$\displaystyle \: \: \frac{a_1}{a_2} \ne \frac{b_1}{b_2}$

CALCULATION :

Given pair of linear equations

$3x + 2y - 4 = 0 \: \: \: and \: \: kx - y - 3 = 0$

Comparing with

$\displaystyle \: a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0$

We get

$\displaystyle \: a_1 = 3 \: , \: b_1 = 2\: , c_1= - 4 \: and \: \: a_2 = k \: , \: b_2 = - 1\: , \: \: c_2= - 3$

So by the given condition

$\displaystyle \: \: \frac{a_1}{a_2} \ne \frac{b_1}{b_2}$

$\implies \: \displaystyle \: \: \frac{3}{k} \ne \frac{2}{ - 1}$

$\implies \: \displaystyle 2k \ne \: - 3$

$\implies \: \displaystyle \:k \ne \: - \frac{3}{2}$

RESULT

$\boxed{\: \: \displaystyle \:k \ne \: - \frac{3}{2} \: \: }$