One equation of a pair of dependent linear equations is 5x + 7y = 2 then the second equation can be

`10x + 14 y + 4 = 0``-10x - 14y +4 =0``-10 +14y +4 =0``10x - 14y + 4 = 0`

Answer : D

Solution : Condition for dependent linear equations
`(a_(1))/(a_(2)) = (b_(1))/(b_(2)) = c_(1)/(c_(2)) = (1)/(k)`
Give equation of line is, `-5x + 7y - 2 = 0`
Here, `" " a_(1) = -5, b_(1) = 7, c_(1) = -2`
From Eq. (i), `" " -(5)/(a_(2)) = (7)/(b_(2)) = - (2)/(c_(2)) = (1)/(k) " " ` [say]
`rArr " " a_(2) = -5k, b_(2) = 7k, c_(2) = -2k`
where, k is any arbitrary constant.
Putting k = 2, then `" " a_(2) = -10 , b_(2) = 14`
and `" " c_(2) = -4`
`:.` The required equation of line becomes
`" " a_(2) x + b_(2) y + c_(2) = 0`
`rArr " " -10x + 14y - 4 = 0`
`rArr " " 10x - 14y + 4 = 0`

Question 19 - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ] - Solutions of Sample Papers for Class 10 Boards

Last updated at Sept. 6, 2021 by

One equation of a pair of dependent linear equations is –5x + 7y = 2. The second equation can be

(a) 10x + 14y + 4 = 0   (b) –10x – 14y + 4 = 0

(c) –10x + 14y + 4 = 0   (d) 10x – 14y = –4

This question is inspired from Question 11 - MCQs from NCERT Exemplar - Chapter 3 Class 10 - Pair of Linear Equations in Two Variables

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Question 19 One equation of a pair of dependent linear equations is –5x + 7y = 2. The second equation can be (a) 10x + 14y + 4 = 0 (b) –10x – 14y + 4 = 0 (c) –10x + 14y + 4 = 0 (d) 10x – 14y = –4 Given equation −5x + 7y − 2 = 0 Therefore, a1 = −5, b1 = 7 , c1 = –2 Since we want a dependent line, it means the lines are coincident For Coincident lines 𝒂𝟏/𝒂𝟐 = 𝒃𝟏/𝒃𝟐 = 𝒄𝟏/𝒄𝟐 Since, a1 = −5 , b1 = 7 , c1 = −2 a2, b2, c2 can be a2 = −10 , b2 = 14 , c2 = −4 Thus, a coincident line is −10x + 14y − 4 = 0 10x − 14y = −4 So, the correct answer is (D)

D. 10x - 14y + 4 = 0

Condition for dependent linear equations -

a1 /a2 = b1/b2 = c1/c2 …(i)

Given equation of line is, - 5x + 7y - 2 = 0;

Comparing with ax+ by +c = 0;

Here, a1 = - 5, b1 = 7, c1 = - 2;

For second equation, let’s assume a2x + b2y + c2 = 0;

From Eq. (i), -5/a2 = 7/b2 = -2/c2 = 1/k

Where, k is any arbitrary constant.

Putting k = - 1/2 then

a2 = 10, b2 = - 14, c2 = 4;

∴ The required equation of line becomes

a2x + b2y + c2 = 0;

10x - 14y + 4 = 0;

asked Sep 6, 2021 in Mathematics by (35.3k points)
closed Sep 7, 2021 by Adarsh01

One equation of a pair of dependent linear equations is –5x + 7y = 2. The second equation can be 

(a) 10x+14y +4 = 0

(b) –10x –14y+ 4 = 0

(c) –10x+14y + 4 = 0

(d) 10x – 14y = –4

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