★ <u>Option (A) y = -5x</u> is the right answer.
<h2><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u></h2><h3>I hope this helps! :)</h3>Answer:
The diagonal is increasing at the rate of 119/104cm/min of the given rectangle.
Step-by-step explanation:
Dimensions of the rectangle
Height = 5cm
Rate of base = 3/2 cm/min
Area = 60cm^2
We know the area of a rectangle of given by = base* Height
b*h = 60
b*5 = 60
b = 12cm
Applying Pythagoras theorem while drawing a diagonal to the rectangle
so our diagonal will be 13cm
Upon differentiating the area of the rectangle we get
b*h = A=60cm^2
using the chain rule of differentiation
h*db/dt + b*dh/dt = 0
b*dh/dt = -h*db/dt
12*dh/dt = -5*3/2
dh/dt = -5/8 cm//min
so the height of the rectangle is decreasing at the rate of -5/8cm/min
now we have all the measurements we need
b = 12 , db/dt = 3/2cm/min
h = 5 , dh/dt = -5/8 cm/min
Upon differentiating we get
2b*db/dt + 2h*dh/dt = 2D*dD/dt
b*db/dt + h*dh/dt = D*dD/dt
12*3/2 + 5*(-5/8) = 13*dD/dt
18 -25/8 = 13*dD/dt
= 13*dD/dt
dD/dt =
Therefore the diagonal is increasing at the rate of 119/104cm/min of the given rectangle.
It is given that two vertices of square are (0,0) and (4,2).
Now the problem is that you haven't given that whether these two vertices are adjacent vertices or opposite vertices of the square.
1. By Supposing that these two are adjacent vertices of Square
The third vertex will be at (-4,2) which lies in third quadrant.
Suppose the coordinate of fourth vertex be (x,y).
Mid point of line joining (4,2) and (-4,2) is{ [4+(-4)]/2,(2+2)/2} is (0,2).
Mid point of line joining (x,y) and (0,0) is (x/2,y/2).
Since diagonals of square bisect each other,
∵ x/2=0
⇒x=0
and
y/2=2
⇒y=4
So, The Coordinate of fourth vertex is (0,4).
Now coming back to second condition if these are two opposite vertex of Square.
Let the third coordinate be (a,b).
Length of diagonal=
Now,let side of Square be A.
Then length of Diagonal of square =√2 A
⇒√2 A=2√5
⇒A =√10
As third vertex is (a,b).
Using distance formula
a² + b²=10 -------------(1)
(a-4)²+ (b-2)²=10 --------------(2)
Solving expression (1) and (2), we get
⇒a²+ b²=(a-4)² +(b-2)²
⇒2a + b =5
⇒b=5-2a
Putting the value of b in (1),we get
⇒a² +(5-2a)²=10
⇒a²+25+4a²-20a =10
⇒5a²-20a+15=0
⇒a² - 4a + 3=0
Splitting the middle term,we get
⇒(a-3)(a-1)=0
⇒a=3 ∧ a=1
we get b=5-2×1=3 and b=5-2×3=5-6=-1
So,the other vertex are (1,3) and(3,-1).
