let two opposite vertices of a rectangle be( 2,1)and( 5,5) if two other vertices lie in y=3 find the equation of sides of rectangle
Answers
Step-by-step explanation:
Since, diagonals of rectangle bisect each other, so mid point of (1,3) and (5,1) must satisfy y=2x+c i.e (3,2) lies on it
⇒2=6+c⇒c=−4
Therefore other two vertices lies on y=2x−4
Let the coordinate of B be (x,2x−4)
Therefore slope of AB . slope of BC=−1
⇒(
x−1
2x−4−3
).(
x−5
2x−4−1
)=−1
⇒(x
2
−6x+8)=0⇒x=4,2⇒y=4,0
Hence, required points are (4,4),(2,0)
Answer:
given, other two coordinates are on line y=3
let coordinate of one vertex be (x,3)
slope of (x,3)and (2,1) = (3-1)/(x-2)
slope of (x,3) and (5,5) = (3-5)/(x-5)
since the shape is rectangle, the slope must be perpendicular to each other
i.e. 2/(x-2/*(-2)/(x-5) = -1
(x-2)x-5) = 4
x^2 - 7x +14 -4 = 0
x^2 - 7x + 10 = 0
this gives x = 5 and 2
thus (5,3) and (2,3) must be the coordinates of other two points
now we can find equation of lines of all the sides
(5,3) and (2,1)
[y2-y1 = (y2-y1)/(x2-x1)*(x-x1)]
y-1 = (3-1)/(5-2)*(x-2)
y-1 = 2/3(x-2)
3y-3 = 2x-4
3y= 2x -1 or 3y - 2x + 1 = 0
in the same way we can find eqn of line of other sides also