If a straight line perpendicular to $2x - 3y + 7= 0$ forms a triangle with the co-ordinate axes whose area is $3$ sq.units, then what is the equation of the straight line?
To find Straight line perpendicular to the given line equation and area with the co-ordinate axes
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geometry
2 Answers
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The slope of that perpendicular line should be $\frac {-3}{2} $ and thus the equation of that line will be $$y = -1.5x +c \Rightarrow 2y+3x =2c =k$$ where $k$ is some constant.
The triangle thus formed will have vertices of $A\,(0,0)\,; B\,(k/3,0)\,;C\,(0,k/2) $. Hope you can take it from here.
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The original line $$ 2x - 3y + 7 = 0 \iff (2,-3) (x,y)^T = -7 $$ has a normal vector $$ n = (2,-3)^T $$ So a perpendicular line has the general equation $$ (3,2) \cdot (x,y)^T = a $$ for some constant $a$, which we can write as $$ y = -(3/2) x + a/2 $$ Then the triangle has the vertices $(0,0)^T$, $(0,a/2)^T$ and $(a/3, 0)^T$
The area of the triangle is $$ A = (1/2) (a/2)(a/3) = a^2 / 12 = 3 $$ so we get $a = \sqrt{36} = 6$.
This gives $$ 3x + 2y = 6 $$ as equation.