0
$\begingroup$

Given two lines $3x = 4y +7$ and $5y =12x + 6$. Find the angle between them and also the equations to the two striaght lines which pass through point $(4,5)$ and make equal angles with two lines.

I can get the angles easily as $\pi - \tan^{-1}(\frac{33}{56})$ and $\tan^{-1}(\frac{33}{56})$, But the problem is the second part.

I know that if a point is given to me as $(x^{\prime} , y^\prime)$ and a line $y =mx + c$, then if we consider the line as the base of a isosceles triangle(it is the unequal side of that isosceles triangle) and the given point is the opposite to the line, then the slope of other two sides are $$ m^\prime = {\tan \alpha \pm m \over 1 \mp m\tan\alpha }$$ where $\alpha$ is equal angle of the triangle.

So to solve that question I supposed that the given two lines are the two equal sides of a isosceles triangle. With $\displaystyle\tan \alpha = \tan\left({\pi - \tan^{-1}(\frac{33}{56}) \over 2}\right), \tan\left({\tan^{-1}(\frac{33}{56}) \over 2}\right) = {11\over 3}, {3\over11}$

Since I know $m^\prime$ in this situation (I assume it to be ${12\over5}$ and ${3\over 4}$(Slopes of the given lines)).

so putting all this in the equation

$$ \left({12\over5} \text{ or } {3\over 4}\right)= {\left({11\over 3} \text { or } {3\over11}\right) \pm m \over 1 \mp m\left({11\over 3} \text { or } {3\over11}\right) }$$

Solving for $m$ in the above equation gives $6$ different values of $m$ which directly implies $6$ different striaght line equations; Of which only two are correct.

  • How can I find these two out from these $6$ equations ?
  • Any easier way of doing this question ?

Here dotted lines are the desired equations and solid lines are the given equations.

(Click on the picture for the link to graphing calculator)

2 Answers 2

1

Let $y=m(x-4)+5$ be the equation of a generic line through $P=(4,5)$. Compute the intersection points $A$ and $B$ of that line with the two given lines (as a function of $m$). Find the squared distances of $A$ and $B$ from $O=(-59/33,-34/11)$, the common point of the given lines. Finally, set $AO^2=BO^2$ to find the values of $m$ which give an isosceles triangle.

  • 0
    How did you get that magic number for O. Is it the intersection of given lines ? – 2017-01-29
  • 0
    Yes, of course. – 2017-01-29
  • 0
    Thanks but can you tell me how to solve this question if we want to do it by my way ? – 2017-01-29
  • 0
    How to eliminate those 6 equations to get the proper two equations ? – 2017-01-29
  • 0
    I'm not sure your equation for $m'$ is correct: letting there $m=0$ (horizontal base) you get the single value $m'=\tan\alpha$. Moreover, what do you mean by "equal angle of the triangle"? – 2017-01-29
  • 0
    Oh sorry. "Equal angle" means that the angle in a isosceles triangle which is opposite to the equal sides, the equal angles in a isosceles triangle. If you are not sure of my formula then I am willing to prove it, if you are willing to help me get an answer to the question "How to eliminate those 6 equations to get the proper two equations ?" – 2017-01-29
  • 0
    I think your formula should be $(m\pm\tan\alpha)/(1\mp m\tan\alpha)$. – 2017-01-29
  • 0
    Yes but isn't it the same ? – 2017-01-29
  • 0
    Not at all: this one also works for $m=0$. – 2017-01-29
  • 0
    The only problem with your method is you have to couple values the right way, but it is not difficult to do that. Take for instance the case $\tan\alpha=11/3$: with $m'=12/5$ you get the solutions $m=-19/147$ and $m=-7/9$, but of course only one of them is right. If now you take instead $m'=3/4$, you get $m=-53/21$ and $m=-7/9$. It is clear then that $m=-7/9$ is the right solution, because you get it in both cases. – 2017-01-29
  • 0
    Ok thats how we will get one solution $m = -7/9$ and by putting $\tan \alpha = 3/11$ we will get the other solution by your obvservation . But do you have a sound arugument when you say "x is the right solution, because you get it in both cases". What if in a special case there were more values that appear twice? – 2017-01-29
  • 0
    That only can happen if $\tan\alpha=1$. But in that case the two values you get correspond to perpendicular lines, so they are exactly the two values giving isosceles triangles. – 2017-01-30
-1

I make you clear that I'm writing equation first as

12y = 5x +6 ----(1) instead of 5y =12x +6

And 2nd eqⁿ is 4y = 3x -7 ------(2)

Here m₁ = 5/12, m₂ =3/4.

Let m is slop of desire line, then

y = m(x-4) +5 ------(3)

Now angle between (1)&(3) will be

Θ₁ = |(12m -5)/(12 +5m)|

Similarly angle between (2)&(3) will be

Θ₂ = |(4m -3)/(4 +3m)|

Now since Θ₁=θ₂ given,then

(12m -5)/(12 +5m) = ± (4m -3)/(4 +3m)

After cross multiplication,we get

48m+36m² -20-15m

= ± (48m-36+20m²-15m)

On Taking +ve sign

16m² = -16, which is unacceptable.

On taking -ve sign

96m+56m² -30m-56=0

⇒56m² +66m -56=0

⇒28m² +33m -28=0

⇒m = (-33±65)/56

    =32/56 or -98/56

    =4/7 or -7/4

Thus desired line would be

7y = 4x -11 & 4y = -7x +33.