Abcd is a square. Co-ordinates of a are (4, 0 ) and coordinates of b are (0, 3). The square lies in the 1st quadrant. Find the co-ordinates of the point of intersection of the diagonals of the square.
Answers
Answer:
7/2 , 7/2
Step-by-step explanation:
Abcd is a square. Co-ordinates of a are (4, 0 ) and coordinates of b are (0, 3). The square lies in the 1st quadrant. Find the co-ordinates of the point of intersection of the diagonals of the square.
a(4,0) b(0,3)
Length of ab = √(0-4)² + (3-0)² = 5
Slope of ab = (3-0)/(0-4) = -3/4
Slope of ad & bc = m
m(-3/4) = -1
=> m = 4/3
y = 4x/3 + c
for ad , it satisfies a(4,0)
0 = 4*4/3 +c
c = -16/3
y = 4x/3 - 16/3
y = 4(x-4)/3
co-ordinates of d = x , 4(x-4)/3
Length of ad = 5 ( as square)
5² = (x-4)² + (4(x-4)/3)²
=> 9 * 25 = 9*(x-4)² + 16(x-4)²
=> 9* 25 = 25(x-4)²
=> (x-4)² = 9
=> x - 4 = ±3
=> x = 1 or 7]
for x=1 , y = 4(1-4)/3 = -4 ( not possible as square lies in 1st quadrant)
y = 4(7-4)/3 = 4
d = (7,4)
b = (0,3)
Diagonal of square intersects at midpoints
mid point of bd = (0+7)/2 , (4 + 3)/2
= (7/2 , 7/2)
similarly we could solve it using
for bc , it satisfies b(0,3)
3 = 4*0/3 +c
c = 3
y = 4x/3 + 3
co-ordinates of c = x , (4x/3 + 3)
Length of bc = 5 ( as square)
5² = (x-0)² + (4x/3 + 3 - 3)²
=> 25 = x² + 16x²/9
=> 25 = 25x²/9
=> x² = 9
=> x = 3 ( as x can not be - 3 , in 1st quadrant)
y = 4*3/3 + 3
y = 7
Co-ordinates of c = (3,7)
a(4,0)
midpoint of ac = 7/2 , 7/2