Question

Q13. Linear relation between x and y such that P(x, y) is equidistant from
the points A (1,4) and B (-1, 2)
(a) X + Y = 3
(b) x-y = 3
(c) x y=3 (d) x/y = 3​

Answer 1

Answer:

Let the points be P(x,y),A(1,4),B(−1,2)

Point P is equidistant from A&B

Now, using the distance formula

x

1

=x,y

1

=y

x

2

=1,y

2

=4

PA=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

=

(1−x)

2

+(4−y)

2

Similarly for PB

x

1

=x,y

1

=y

x

2

=−1,y

2

=2

PA=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

=

(−1−x)

2

+(2−y)

2

we know that

PA=PB

(1−x)

2

+(4−y)

2

=

(−1−x)

2

+(2−y)

2

Squaring both sides

⇒(1−x)

2

+(4−y)

2

=(−1−x)

2

+(2−y)

2

⇒1+x

2

−2x+16+y

2

−8y=1+x

2

+2x+4+y

2

−4y

⇒−2x−2x−8y+4y+16−4=0

⇒−4x−4y+12=0

⇒x+y−3=0

heyyy!!! please mark my answer brainliest

Answer 2

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Let the points be P(x,y),A(1,4),B(−1,2)

Point P is equidistant from A&B

Now, using the distance formula

x

1

=x,y

1

=y

x

2

=1,y

2

=4

PA=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

=

(1−x)

2

+(4−y)

2

Similarly for PB

x

1

=x,y

1

=y

x

2

=−1,y

2

=2

PA=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

=

(−1−x)

2

+(2−y)

2

we know that

PA=PB

(1−x)

2

+(4−y)

2

=

(−1−x)

2

+(2−y)

2

Squaring both sides

⇒(1−x)

2

+(4−y)

2

=(−1−x)

2

+(2−y)

2

⇒1+x

2

−2x+16+y

2

−8y=1+x

2

+2x+4+y

2

−4y

-2x−2x−8y+4y+16−4=0

⇒−4x−4y+12=0

⇒x+y−3=0

$\huge\mathcal{\fbox{\fbox{\pink{Hope It helps you..}}}}$