Question

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

Answer 1

Question:

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

To find:

★ To find the points .

Answer:

$The \: points \: are \: \underline{ \: \sf \pink{ ( \frac{15}{7} , \frac{ - 5}{14})} \: }$

Given:

★ An equation of a straight line is given, 3x + 4y = 5.

It also can be written as, 3x + 4y - 5 = 0.

★ Two points are given, ( 1,2 ) and (3,4).

Step-by-step explanation:

Let the point on the line 3x + 4y - 5 = 0 equidistant from points A ( 1, 2 ) and B ( 3 , 4 ) be C ( x , y )

$3x + 4y - 5 = 0 ---> (1)$

We know that, Distance formula

$\boxed{d = \sqrt{ ({x _{2} - x _{1} )}^{2} + {(y _{2} - y _{1} ) }^{2} } }$

$\sqrt{ {(x - 1)}^{2} + {(y - 2)}^{2} } = \sqrt{ {(x - 3)}^{2} + {(y - 4)}^{2} }$

Square root get cancelled.

$\sf{(x - 1)}^{2} + {(y - 2)}^{2} = {(x - 3)}^{2} + {(y - 4)}^{2}$

Now implying, ( a - b ) ² formula.

$\sf {x}^{2} - 2x + 1 + {y}^{2} - 4y + 4 = {x}^{2} - 6x + 9 + {y}^{2} - 8y + 16$

$\sf {x}^{2} + {y}^{2} - 2x - 4y + 5 = {x}^{2} + {y}^{2} - 6x - 8y + 25$

$\sf \bold{{x}^{2} + {y}^{2}} - 2x - 4y + 5 \bold{ - {x}^{2} - {y}^{2}} + 6x + 8y - 25 = 0$

$\sf \boxed{4x + 4y - 20= 0}---> (2)$

Now , subtracting ( 1 ) and ( 2 )

3x + 4y - 5 = 0

4x + 4y - 20 = 0

( - ) ( - ) ( + )

_____________

7x - 15 = 0

_____________

$\hookrightarrow\sf 7x = 15$

$\hookrightarrow \sf \boxed{x = \frac{15}{7} }$

Now substituting the value of x in ( 1 )

$\mapsto \sf3 \times \frac{15}{7} + 4y - 5 = 0$

$\mapsto\sf 45 + 28y - 35 = 0$

$\mapsto\sf 28y + 10 = 0$

$\mapsto \sf28y = - 10$

$\mapsto \sf14y = - 5$

$\mapsto \sf \boxed{y = \frac{ - 5}{14} }$

$\therefore$The points are $\bold{(\frac{15}{7} , \frac{ - 5}{14}) }$

Formula used:

$\bigstar d = \sqrt{ ({x _{2} - x _{1} ) }^{2} + {(y _{2} - y _{1} ) }^{2} }$

$\bigstar {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2}$