The straight lines `4a x+3b y+c=0` , where `a+b+c` `(4,3)` (b) `(1/4,1/3)` `(1/2,1/3)` (d) none of these

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The straight lines `4a x+3b y+c=0` , where `a+b+c` `(4,3)` (b) `(1/4,1/3)` `(1/2,1/3)` (d) none of these

The straight lines `4a x+3b y+c=0` , where `a+b+c` `(4,3)` (b) `(1/4,1/3)` `(1/2,1/3)` (d) none of these
A. (4,3)
B. (1/4,1/3)
C. (1/2,1/3)
D. none of these

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
The set of lines is 4ax+3by+c=0, where a+b+c=0.
Eliminating c, we get
4ax+3by-(a+b)=0
a(4x-1)+b(3y-1)=0
This passes through the intersection of the lines 4x-1=0 and 3y-1=0, i.e., x 1/4, y=1/3, i.e.,(1/4, 1/3).

4 views Answered 2023-02-05 15:29