Question

Determine the coordinates of a point which is equidistant from the point (1,2) and (3,4) and the shortest distance from the line joining the points (1,2) and (3,4) to the required point is √2 .​

Answer 1

Answer:

is equidistant from the point (1,2) and (3,4) and the

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Answer 2

$\large{\boxed{\boxed{\sf{ Question}}}}$

Determine the coordinates of a point which is equidistant from the point (1,2) and (3,4) and the shortest distance from the line joining the points (1,2) and (3,4) to the required point is √2 .

$\large{\boxed{\boxed{\sf{According \: To \: The \: Question}}}}$

We had to Determine the coordinates of a point

•who is equidistant from the point (1,2) and (3,4)

• the shortest distance from the line joining the points (1,2) and (3,4)

• the required point is √2 .

$\large{\boxed{\boxed{\sf{Solution}}}}$

Let the point be A(1, 2) and B(3, 4)

The mid-point of the line joining A and B is C (2,3)

Slope of line AB =

$\sf \dfrac{4 - 2}{3 - 1} = 1$

Let the required point be

D $\sf( \alpha , \beta )$

Then D must be a point on the line perpendicular to the line AB and passing through point C

∴ Slope of CD = - 1

Equation of CD

y - 3 = - 1(x - 2)

x + y = 5

Equation of AB

y - 2 = 1(x - 1)

x - y + 1 = 0

The point

D $\sf( \alpha , \beta )$ must satisfy the equation

x + y = 5

$\sf∴ \alpha + \beta = 5.....(1)$

The perpendicular distance from $\sf( \alpha , \beta )$ to AB is

$\sf \dfrac{ \alpha - \beta + 1}{ \sqrt{2} } = \sqrt{2}$

$\sf∴ \alpha - \beta = 1.....(2)$

Solving equations 1 and 2

$\sf \alpha = 3, \beta = 2$