Asked by nuru
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
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
Answers
Answered by
Damon
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
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
Answered by
Steve
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
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
Answered by
Anonymous
Where did the √2^2+1^2 come from
Try to break it down please it's confusing
Try to break it down please it's confusing
Answered by
Anonymous
How did you get the 13
It's giving me -78/6
It's giving me -78/6
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