Question

The point in the line 3x+4y=5 which is equidistant from (1,2) and (3,4)

Answer 1

Answer:

Let's take a general point on the line 3x+4y−5=0, which is (x,

4

5−3x

)

According to given condition, this point is equidistant from points (1,2) and (3,4)

Using distance formula, by equalizing their distances we get,

⇒

(2−

4

5−3x

)

2

+(1−x)

2

=

(4−

4

5−3x

)

2

+(3−x)

2

⇒(2)

2

+(

4

5−3x

)

2

−(5−3x)+1

2

+x

2

−2x=(4)

2

+(

4

5−3x

)

2

−2(5−3x)+3

2

+x

2

−6x

⇒x=15

⇒ Hecne y=

4

5−3(15)

=−10

So the point is (15,−10)

Step-by-step explanation:

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