Question

the point on the line 3x +4y =5 which is equidistance from (1,2)and (3,4) is

Answer 1

Answer:

( 15, - 10 )

Step-by-step-explaination:

Let that point be ( a, b ).

As ( a, b ) lies on 3x + 4y = 5,

= > 3(a) + 4(b) = 5

= > 3a + 4b = 5

= > 5 - 3a = 4b ... (1)

As it is equidistant from the given points.

= > distance b/w ( 1, 2 ) and ( a, b ) = distance b/w ( 3, 4 ) and ( a, b )

Using distance formula,

= > √{ ( 1 - a )² + ( 2 - b )² } = √{ ( 3 - a )² + ( 4 - b )² }

= > ( 1 - a )² + ( 2 - b )² = ( 3 - a )² + ( 4 - b )^2

= > 1 + a² - 2a + 4 + b² - 4b = 9 + a² - 6a + 16 + b² - 8b

= > 4a + 4b = 20

= > 4a - 3a + 5 = 20 { 4b = 5 - 3a }

= > a = 15

Hence, 4b = 5 - 3(15) = - 40

b = - 10

Hence the required point is ( 15, - 10 ).