State the equations of some straight lines

Question

$21x+8y=18$ and $11x+3y+12=0$ are two straight lines.

1. Find the intersecting point.
2. State the equations of the straight lines which pass through the intersecting point and make a triangle with the x and y axis with an area 9 unit square.

I have tried. By solving this equations I got the intersecting point A =(-6,18). I assumed the gradient of the straight line which pass through A point and make a triangle of area 9 unit square is m. So, the equation of the straight line is $$y-18=m(x+6)$$ =>$$y-18=mx+6m$$ =>$$mx-y=-6m-18$$ =>$$x/-6m-18/m+y/6m+18=1$$ So, the straight line intersect $x$ axis at $(-6m-18/m,0)$ and intersect y axis at $(0,6m+18)$ So the area of the triangle is $=(1/2)×\frac{(6m+18)}{m}×(6m+18)$

So, $$ \frac{1}{2}\frac{(6m+18)}{m}×(6m+18)=9$$ By solving this equation I got two value of m. But these are imaginary. So, my answer is not same with the answer given in the book. Any Idea?

Answer 1

multiplying the first equation b $-3$ and the second one by $8$ we get $$-63x-24y=-54$$ and $$88x+24y=-96$$ by adding both we obtain $$25x=-150$$ can you proceed?

Answer 2

The two lines have homogeneous coordinates $L_1 = (21,8,-18)$ and $L_2 = (11,3,12)$. Their intersection point $P$ has homogeneous coordinates:

$$ P = \begin{pmatrix} 21 \\ 8 \\ -18 \end{pmatrix} \times \begin{pmatrix} 11 \\ 3 \\ 12 \end{pmatrix} = 25 \begin{pmatrix} 6 \\ -18 \\1 \end{pmatrix} $$

which means the location of $P$ is $(6,-18)$.

_{NOTE: $\times$ is the vector cross product}

This I have confirmed with geogebra.

Now a generic line $M: (-\sin \psi)x + (\cos \psi) y - d = 0 )$ interseects the point $P$ when $ d= -18 \cos\psi - 6 \sin \psi$.

_{NOTE: The line direction is $(\cos\psi,\sin\psi)$ and the distance from the origin is $d$.}

So the equation of the line is $$(y-18) \cos\psi - (x+6) \sin \psi = 0$$

This line intersects the x and y axes at

$$ \begin{aligned} x_c & = -18 \cot \psi -6 \\ y_c & = 6 \tan \psi + 18 \end{aligned} $$

The area described in the problem is $A = \frac{1}{2} x_c y_c$ which gives the following equation

$$ 9 = -162 \cot \psi - 18 \tan \psi - 108 $$

Sbstitute $z = \tan \psi$ and solve $ 117 = -162 \frac{1}{z} -18 z$ for

$$ z = \begin{cases} -2 \\ -\frac{9}{2} \end{cases} $$

$$ \psi = \tan^{-1}(z) = \begin{cases} \tan^{-1} \left( \frac{1}{2} \right) - \frac{\pi}{2}\\ \tan^{-1} \left( \frac{2}{9} \right) - \frac{\pi}{2} \end{cases} $$

The two possible solutions give the following equations of a line

$$\begin{cases} (y-18) \frac{1}{\sqrt{5} } + (x+6) \frac{2}{\sqrt{5} } =0 \\ (y-18) \frac{2}{\sqrt{85} } + (x+6) \frac{9}{\sqrt{85} } = 0 \end{cases} $$

Both lines give an area of 9 (you should check), but each lies on a different side of the origin.

A visual check is done again with geogebra (yellow line is the first solution and green the second).

Answer 3

You have some errors in your notation: you wrote $−6m−18/m$, which is $−6m−(18/m)$, when really you meant $(−6m−18)/m$.

But it is clear enough that you correctly understood where the axis intercepts are: \begin{align} \text{$x$ intercept: }& \left(\frac{-6m-18}{m}, 0 \right) \\ \text{$y$ intercept: }& \left( 0, 6m + 18 \right) \\ \end{align}

Your mistake was to take one leg of the triangle equal to $\dfrac{6m+18}{m}$ and the other equal to $6m+18$.

If $m$ is positive then the triangle contains the rectangle with vertices $(-6,0)$, $(-6,18)$, $(18,0)$, and $(0,0)$, whose area is much larger than $9$. So $m$ must be negative.

If $6m+18 > 0$ then $\dfrac{-6m-18}{m} > 0$ (negative divided by negative) and you have one leg of length $6m+18$ and the other of length $\dfrac{-6m-18}{m}$.

If $6m+18 < 0$ then $\dfrac{-6m-18}{m} < 0$ (positive divided by negative) and you have one leg of length $-(6m+18)$ and the other of length $-\dfrac{-6m-18}{m} = \dfrac{6m+18}{m}$.

In either case, the area of the triangle is $$ -\frac12 \left(6m+18 \right)\left(\dfrac{6m+18}{m} \right) = 9. $$ The only change between your equation and this one is the negative sign on the left side of the equation, but that is enough to make the equation have real solutions.