Question

Find the distance between parallel lines.
4x + 3y = 11 and 8x + 6y = 15​

Answer 1

Step-by-step explanation:

Given−

theequationsoftwoparallellinesare

4x+3y=11⟹4x+3y−11=0

≡a

1

x+b

1

y+c

1

=0 .......(i)and

8x+6y=15⟹8x+6y−15=0

≡a

2

x+b

2

y+c

2

=0 ........(ii).

Tofindout−

theperpendiculardistancebetweentwoparallellines=?

Solution−

Theperpendiculardistancedbetweentwoparallellines

=differencedbetweentheirdistancesfromtheorigin.

Nowtheperpendiculardistance=pofalineax+by+c=0

fromtheoriginisp=

a

2

+b

2

c

.

From(i)a

1

=4,b

1

=3&c

1

=−11.

∴p

1

=

a

1

2

+b

1

2

c

1

=

∣

∣

∣

∣

∣

4

2

+3

2

−11

∣

∣

∣

∣

∣

units=

5

11

units.

Again,from(ii),a

2

=8,b

2

=6&c

1

=−15.

∴p

2

=

a

2

2

+b

2

2

c

1

=

∣

∣

∣

∣

∣

8

2

+6

2

−15

∣

∣

∣

∣

∣

units=

10

15

units.

Sod=∣p

1

−p

2

∣=

∣

∣

∣

∣

∣

5

11

−

10

15

∣

∣

∣

∣

∣

units=

10

7

units.

∴Thedistancebetweenthegivenlines=

10

7

units.

Answer 2

question ☟☟

• Find the distance between parallel lines.

4x + 3y = 11 and 8x + 6y = 15

solution☟☟

Given lines are 4x + 3y = 11 and 4x + 3y = 15/2

Distane between two parallel lines

|c₁ - c₂ / √a² + b²|

= | 11 - / √16 + 9 |

= | 7 / 2 × 5 |

=7/10

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