Question

one vertex of a square ABCD is A(-1,1) and the equation of one diagonal BD is 3x+y-8=0 then C=​

Answer 1

Answer:

The coordinate of C is (5, 3)

Step-by-step explanation:

Let the coordinates of C are (x', y')

Since AD will be the another diagonal, the mid point of A and D will lie on the diagonal BD

i.e. it will satisfy the equation of BD

The mid point of A and D is $(\frac{x'-1}{2}, \frac{y'+1}{2})$

this will lie on the line 3x + y - 8 = 0

Thus,

$3(\frac{x'-1}{2})+\frac{y'+1}{2}=8$

$\implies 3x'-3+y'+1=16$

$\implies 3x'+y'=18$   ..........................(1)

Also, both the diagonals will be perpendicular

Slope of AC = $\frac{y'-1}{x'+1}$

Slope of BD = -3

Thus

$-3\times \frac{y'-1}{x'+1}=-1$

$\implies 3y'-3=x'+1$

$\implies x'-3y'=-4$   ................... (2)

Multiplying eq (1) by 3 and adding it to eq (2)

10x' = 50

or, x' = 5

Putting the value of x' in eq (1)

3×5 + y' = 18

or, y' = 18 - 15 = 3

Thus the coordinate of C is (5, 3)