Question

Abcd is a square a=(1 2) b=(3 -4).If line cd passes through (3,8) then mid point of cd os

Answer 1

Coordinate Geometry

Solution:

The coordinates of A and B are (1, 2) and (3, - 4) respectively.

The equation of the side AB is

(y - 2)/(2 + 4) = (x - 1)/(1 - 3)

or, (y - 2)/6 = (x - 1)/(- 2)

or, (y - 2)/3 = - (x - 1)

or, y - 2 = - 3x + 3

or, 3x + y = 5

Since the side CD is parallel to the side AB, let the equation of AB be

3x + y = k

Given, the side CD passes through the point (3, 8),

3 (3) + 8 = k

or, k = 17

Thus the the equation of CD is

3x + y = 17

The mid-point of AB is

( (1 + 3)/2, (2 - 4)/2 )

i.e., (2, - 1)

The straight line perpendicular to the mid-point of the side CD is given by

x - 3y = k, which passes through the point (2, - 1)

Thus 2 + 3 = k, i.e., k = 5

Then the equation of the straight line perpendicular to the mid-point of the side CD is

x - 3y = 5

NOW, we solve the equations 3x + y = 17 and x - 3y = 5 to find the mid-point of CD.

3x + y = 17 .....(1)

x - 3y = 5 .....(2)

Eliminating x from (1), (2), we get

3 (3y + 5) + y = 17

or, 9y + 15 + y = 17

or, 10y = 2

or, y = 1/5

Putting y = 1/5 in (2), we get

x - 3/5 = 5

or, x = 5 + 3/5

or, x = 28/5

Answer: The mid-point of CD is (28/5, 1/5).