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Q.

Find the joint equation of the lines which pass through the point (4, −3) and are parallel to the lines x=1 and y=5.


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a

3x − y + xy – 12 = 0

b

3x − 4y + xy2 – 12 = 0

c

3x − 4y + xy – 12 = 0

d

None of these 

answer is C.

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Detailed Solution

The line parallel to the line x=1 will be of the form x=c
Now, x = c passes through the point (4, −3)
So, 4 = c
And the line is x = 4
x−4=0 ........... (i)
The line parallel to the line y = 5 will be of the form y = k
Now, y = k passes through the point (4, −3)
So, −3 = k
And the line is y = −3
y + 3 = 0 ........... (ii)
Joint equation of the lines is,
(x−4) (y+3) = 0
3x − 4y + xy – 12 = 0
Hence, option 3 is correct.
 
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