Question

for which value(s) of p, will the lines represented by the following pair of linear equation be parallel 3x-y-5=0 6x-2y-p=0​

Answer 1

since, the lines are parallel.......

here's your result

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Answer 2

Given: Equations -

• 3x - y - 5 = 0.
• 6x - 2y - p = 0.

To find: The value of p such that the lines so formed by the equation will be parallel.

Answer:

Let's look at the equations -

$\sf a_1x\ +\ b_1y\ + c_1\ =\ 0\ and\ a_2x\ +\ b_2y\ +\ c_2\ =\ 0.$

$\sf If\ \dfrac{a_1}{a_2}\ \ne\ \dfrac{b_1}{b_2}, then\ the\ pair\ of\ equations\ has\ exactly\ 1\ solution; intersecting\ lines.\\\\\\\\If\ \dfrac{a_1}{a_2}\ =\ \dfrac{b_1}{b_2}\ =\ \dfrac{c_1}{c_2},\ then\ the\ pair\ of\ equations\ has\ infinitely\ many\ solutions; coincident\ lines.\\\\\\\\If\ \dfrac{a_1}{a_2}\ =\ \dfrac{b_1}{b_2}\ \ne\ \dfrac{c_1}{c_2}, then\ the\ pair\ of\ equations\ has\ no\ solution; parallel\ lines.$

According to the question, the lines must be parallel. So the last condition will be used.

From the given points, we have,

$\sf a_1\ =\ 3\\\\b_1\ =\ -1\\\\c_1\ =\ -5\\\\a_2\ =\ 6\\\\b_2\ =\ -2\\\\c_2\ =\ -p$

Using the condition,

$\sf \dfrac{3}{6}\ =\ \dfrac{-1}{-2}\ \ne\ \dfrac{-5}{-p}\\\\\\Let's\ take\ the\ second\ and\ the\ last\ fractions.\\\\\\\dfrac{-1}{-2}\ \ne\ \dfrac{-5}{-p}\\\\\\Cross\ multiplying,\\\\\\p\ \ne\ 10$

Therefore, p can be any value but 10.