How do you find the area of a triangle bounded by the y axis, the line f(x) = 7 - 4/5 x, and the line perpendicular to f(x) that passes through the origin?

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Answer 1 · 2018-08-02T12:54:44

Area of the triangle is $11.95$ sq.unit

$f(x)=y= 7- 4/5 x$ , slope is $m=-4/5$

Slope of perpendicular line is $m_p= -1/(-4/5)=5/4$

Equation of perpendicular line passing through origin is

$y=5/4 x$ , intersecting point between the lines is

$5/4 x= 7- 4/5 x or 25 x= 140- 16 x or 41 x = 140$

$:. x = 140/41 , y= 5/4*140/41=175/41$

y intercept of line is $y= 7- 4/5 x ; y= 7$ , so the triangle is

bounded by the points $(0,0) ,(0,7) and (140/41,175/41)$

Area of the triangle is $A_t=1/2*7*140/41=490/41~~ 11.95$

Area of the triangle is $11.95$ sq.unit [Ans]