Question #287008

find the equation of the line through the point (5,6) which forma with the axes a triangle whose area is 80.


1
Expert's answer
2022-01-16T12:39:11-0500

Let the equation of the line is y=mx+b,m0.y=mx+b, m\not=0.

yy -intercept: x=0,y=bx=0, y=b

Point (0,b).(0, b).

xx -intercept: y=0,0=mx+b=>x=bmy=0, 0=mx+b=>x=-\dfrac{b}{m}

Point (0,b).(0, b).

The area of the triangle


Area=12bbm=80Area=\dfrac{1}{2}|b|\cdot|-\dfrac{b}{m}|=80

b2160=m\dfrac{b^2}{160}=|m|

Line passes through the point (5,6)(5, 6)


6=m(5)+b6=m(5)+b

m=6b5m=\dfrac{6-b}{5}

b<6b<6

b2160=6b5\dfrac{b^2}{160}=\dfrac{6-b}{5}

b2+32b192=0b^2+32b-192=0

(b+16)2=448(b+16)^2=448

b=16±87b=-16\pm8\sqrt{7}

b=1687b=-16-8\sqrt{7}


m=6(1687)5=22+875m=\dfrac{6-(-16-8\sqrt{7})}{5}=\dfrac{22+8\sqrt{7}}{5}

y=22+875x(16+87)y=\dfrac{22+8\sqrt{7}}{5}x-(16+8\sqrt{7})



b=16+87<6b=-16+8\sqrt{7}<6


m=6(16+87)5=22875m=\dfrac{6-(-16+8\sqrt{7})}{5}=\dfrac{22-8\sqrt{7}}{5}

y=22875x+(16+87)y=\dfrac{22-8\sqrt{7}}{5}x+(-16+8\sqrt{7})


b>6b>6

b2160=6b5\dfrac{b^2}{160}=-\dfrac{6-b}{5}

b232b+192=0b^2-32b+192=0

(b16)2=64(b-16)^2=64

b=16±8b=16\pm8

b=168=8b=16-8=8

m=685=25m=\dfrac{6-8}{5}=-\dfrac{2}{5}

y=25x+8y=-\dfrac{2}{5}x+8

b=16+8=24b=16+8=24

m=6245=185m=\dfrac{6-24}{5}=-\dfrac{18}{5}

y=185x+24y=-\dfrac{18}{5}x+24


Answer:


y=22+875x(16+87)y=\dfrac{22+8\sqrt{7}}{5}x-(16+8\sqrt{7})

y=22875x+(16+87)y=\dfrac{22-8\sqrt{7}}{5}x+(-16+8\sqrt{7})

y=25x+8y=-\dfrac{2}{5}x+8

y=185x+24y=-\dfrac{18}{5}x+24


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