The line l is perpendicular to line m. m passes through (-3,7) and (2,-9). l passes through the midpoint of m. find the area of the triangle made between l, y axis and x axis

1.1390625

Step-by-step explanation:

The area will be half the magnitude of the product of the x- and y-intercepts. Those interceps can be found from the equation of the perpendicular bisector of M.

The midpoint of M is ...

... ((-3, 7) + (2, -9))/2 = (-0.5, -1)

From one end of M to the other, the change in x is 2-(-3) = 5. The corresponding change in y is -9-7 = -16. For change values ∆x and ∆y and midpoint (h, k), the perpendicular bisector through (h, k) can be written ...

... ∆x(x -h) +∆y(y -k) = 0

For our numbers, this is

... 5(x +0.5) -16(y +1) = 0

... 5x -16y = 13.5 . . . . . simplified, almost standard form

... 10x -32y = 27 . . . . . in standard form

We can benefit by putting this equation into intercept form. Dividing by 27 does it.

... x/(27/10) -y/(27/32) = 1

The x-intercept is 27/10 = 2.7; the y-intercept is -27/32 = 0.84375. The enclosed area is half the product of the magnitudes of these values:

... Area = (1/2)(27/10)(27/32) = 729/640 = 1.1390625

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In the attachment, the area is rounded to 5 decimal places.

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

or

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.

answer: h

step-by-step explanation: