Question

The linear equation 2x-3y+ 4=0 has

1)a unique solution
2) solutions
3)infinitely many solutions
4)no solution

Answer 1

Answer:

A unique solution

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Answer 2

Answer:

R.E.F image

Perpendicular distance of point (x

1

,y

1

)

from a line ax+by+c=0 is

given by

±d=

a

2

+b

2

ax

1

+by

1

+c

Perpendicular distance (d)=2

Case 1 : for d=±2

equation of line : 3x+4y−6=0

a=3 b=4 c=−6

∴=

3

2

+4

2

3x

1

+4y

1

−6

∴10=3x

1

+4y

1

=6

∴3x

1

+4y

1

−16...(1)

Since the point lies on 2ax+3y+4=0

2x

1

+3y

1

=−4...(2)

2x

1

=−4−3y

1

x

1

=−2

2

−3

y

1

...(2)

Substituting equation (2) in equation (1)

∴3(−2−

2

3

y

1

)+4y

1

=16

∴−6−

2

9

y

1

+4y

1

=16

y

1

=−44

Substituting y

1

=−44 in (2)

∴x

1

=−2−

2

3

×(−44)

∴x

1

=64

∴ first point is of (64,−44)

Case 2 for d=−2

∴−2=

32+42

3x

1

+4y

1

−6

∴−10=3x

1

+4y

1

−6...(3)

Substituting equation (2) in equation (3)

∴3(−2−

2

3

y

1

)+4y

1

=−4

∴−6

2

−9

y

1

+4y

1

=−4

y

1

=−4

Substituting the above in equation (2)

∴x

1

=−2−

2

3

×(−4)

∴x

1

=4

∴2

nd

point is (4,−4)

Step-by-step explanation:

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