Answer to Question #287008 in Analytic Geometry for jay

Question

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

Answer 1

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

$y$ -intercept: $x=0, y=b$

Point $(0, b).$

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

Point $(0, b).$

The area of the triangle

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

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

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

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

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

$$b<6$$

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

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

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

$$b=-16\pm8\sqrt{7}$$

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

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

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

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

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

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

$$b>6$$

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

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

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

$$b=16\pm8$$

$b=16-8=8$

$$m=\dfrac{6-8}{5}=-\dfrac{2}{5}$$

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

$b=16+8=24$

$$m=\dfrac{6-24}{5}=-\dfrac{18}{5}$$

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

Answer:

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

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

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

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