If line 2x+y=-3 and x-2y=1 form the sides of a
rectangle whose other sides meet at the point
(3, 4), find the equations of the other sides and
the area of the rectangle
plz help show solution step
y = -2x -3 the slope = -2
y = (1/2)x - 1/2 slope = 1/2 so perpendicular, whew that helps.
where do they cross? That is a corner.
(x-1)/2 = -2x-3
x-1 = -4x -6
5x = -5
x=-1
y = 2-3 = -1
so other corner is (-1, -1)
now draw it and find the other two intersetions
Where did the √2^2+1^2 come from
Try to break it down please it's confusing
How did you get the 13
It's giving me -78/6
To find the equations of the other sides of the rectangle and calculate its area, we first need to determine the coordinates of the other two vertices of the rectangle.
Given that the two sides meet at the point (3, 4), we can substitute these values into the equations of the given sides and solve for x and y.
For the first side with the equation 2x + y = -3:
Substituting x = 3 and y = 4:
2(3) + 4 = -3
6 + 4 = -3
10 = -3
The equation is not satisfied, which means (3, 4) is not a point on this side. Therefore, the sides x - 2y = 1 and 2x + y = -3 do not form a rectangle.
Double-checking the equations, it seems there might be a mistake in the given problem. Please review the problem statement and equations and provide the correct information, so I can assist you further in finding the solution.
One side is the distance from (3,4) to 2x+y+3=0
The other side is the distance from (3,4) to x-2y-1=0
The area of the rectangle is thus
|2*3+1*4+3)/√(2^2+1^2) * |1*3-2*4-1|/√(1^2+2^2) = 13