Question

One side of a rectangle lies along the line 4x+7y+5=0. two of its verticle are (-3,1) and (1,1). then the eqn of other sides are

Answer 1

Solution :-

The point (3,1) lies on this line 4x+7y+5 = 0

so,

let A (-3,1) and c (1,1) be 2 vertices of the rectangule ABCD

equation of line AB is 4x +7y +5 =0

CD is parallel to AB passing through c (1,1)

let the equation of CD be 4x+7y+k=0 ...........(i)

this passes through c (1,1)

4 (1) +7(1) +k=0
=> k=-11

putting the value of K in equation (i)

equation is the line CD is 4x +7y-11=0

As AD Parallel AB equation of line AD is 7x-4y+14 =0 ........(ii)

this passes through A (-3,1)

7 (-3) -4 (1) + m=0
=> m=25
putting the value of min equation ...(ii)

equation of line AD is 7x-4y+25=0 ........ (iii)

this passes through c (1,1)

7 (1) - 4 (1) +x =0
x= -3

putting the value of x in equation ...... (iii)

equation of line BC is 7x -4y -3 =0



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I hope it's help you.....!!! :)
 

Answer 2

since , side AB perpendicular to AD.
so, its equation is in the form of 7x - 4y + R = 0
since, it passes through (-3, 1)
so, 7 × (-3) - 4 × 1 + R = 0
=> -21 - 4 + R = 0
=> R - 25 = 0
=> R = 25
so, equation of AB is 7x - 4y + 25 = 0

now, BC is parallel to AD, therefore , its equation is 4x + 7y + T = 0
since, it passes through (1,1),
4 × 1 + 7 × 1 + T = 0
=> 4 + 7 + T = 0
=> 11 + T = 0
=> T = -11
so, equation of BC is 4x + 7y - 11 = 0

now, equation of DC is 7x - 4y + Q = 0
since, it passes through (1,1),
7 × 1 - 4 × 1 + Q = 0
=> 7 - 4 + Q = 0
=> 3 + Q = 0
=> Q = -3
so, equation of DC is 7x - 4y - 3 = 0