What are the lengths of lines M to M1 and M to M2 in this geometry problem?

In summary: Basically you multiply the leftmost equation by the rightmost equation and the result is the solution.I once knew how to do it but I have long since forgotten how it is supposed to... Basically you multiply the leftmost equation by the rightmost equation and the result is the solution.I think my teacher used to say that there's about 4 main ways to solve systems of equations.I think that the "equating to each other" method is probably the simplest, and the "summation method" is probably the most versatile.
  • #1
late347
301
15

Homework Statement


line m2-to-m is 3km longer than line m1-to-m

what are lengths for ##M~~M_2## and ##M~~M_1##
geometria kuva kolmio m m1 m2.jpg

Homework Equations


pythagorean theorem hopefully can be used

The Attempt at a Solution



geometria kuva kolmio edistynyt.png

use the picture to your advantage in hopefully creating a valid system of equations. With which, X and Y can both be solved.

However the question only asked really what was (X-3) and what was (X). Y should hopefully only be used as an intermediary variable to that purpose(?)

system of equations is as follows:
$$y^2~+~ 5.6^2~= ~x^2$$
$$(18-y)^2 ~+ ~5.6^2~ = ~(x-3)^2$$
The top equation can be reduced into following form...
##x= \sqrt{y^2+5.6^2}##
the lower equation can be reduced into the form of...
##y^2~-36y~+346.36= x*(x-6)##

insert top equation into the lower equation. Hopefully this was correct because on pen-paper-style it took an entire page, really...
##35y^2 -630y +2724.89=0##

from calculator
y = 10.7736967
y= 7.226303295

We can test these Y's and hopefully to ascertain whether or not the system is true.

I think that with the top value of Y (y ~ 10.773 ...)

the originatl system of equations seems to work ok.

testing can be done with calculator for example with assigning variables with some values such as
y=10.7736967
from the top equation we plug in the y value, and calculate that x should be equal to about 12.14218022with that x value, you can plug it into the lower equation and it should hold true.

so it looks like length of x = 12.14218022 km
it looks like length of (x-3) = 9.142180224 km
 
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  • #2
feedback or critique is welcome. Especially if I was wrong ... (obviously...)

also alternative tactic of solving is welcome.

maybe this is dumb question but how can a person know for a fact. WHICH side from a given triangle with a 90deg angle, is supposedly the hypotenuse of that particular triangle? Without knowing strictly the lengths of sides at first instance?
 
  • #3
There is an easier way. If you take the difference of your two original equations the quadratics cancel, leaving you a linear relationship between x and y.
late347 said:
how can a person know for a fact. WHICH side from a given triangle with a 90deg angle, is supposedly the hypotenuse of that particular triangle? Without knowing strictly the lengths of sides at first instance?
I assume you mean, if you are given two side lengths and the fact that the triangle is right-angled, but don't know the relationship between that and the given lengths. If so, you cannot tell. In a2+b2=c2, you can plug in any two values for a and b and solve, or plug the larger in as c and the smaller in as a and solve.
 
  • #4
Well I googled it and it looks like in a right angle triangle...

Even if you don't know any lengths... then hypotenuse ought to be that length which is opposite to the right angle (the right angle's location you woyld know initially)

I did not understand what you meant by difference of squares though
 
  • #5
late347 said:
it looks like in a right angle triangle..
What does?
late347 said:
hypotenuse ought to be that length which is opposite to the right angle
Of course.
late347 said:
I did not understand what you meant by difference of squares though
In your first equation, you have a y2 on the left and x2 on the right. If you expand the parentheses in the second equation that will also have those squared terms. If you subtract the first equation from the second (left from left, right from right) you will have an equation with no quadratic terms in x and y.
 
  • #6
haruspex said:
What does?

Of course.

In your first equation, you have a y2 on the left and x2 on the right. If you expand the parentheses in the second equation that will also have those squared terms. If you subtract the first equation from the second (left from left, right from right) you will have an equation with no quadratic terms in x and y.

I see now, but yet I wonder whether or not that is really so much easier in actual fact.

I think my teacher used to say that there's about 4 main ways to solve systems of equations.

Placement method (gauss's method??). He never really had time to explain or demonstrate that method in class, because our progress in math class was somewhat at a slow rate. I think we only really had some homework problems with systems of equations. Not those problems in the last exam.

summation method (reduce one of the equations in the pair, into suitable form, and multiply by (-1) and sum the top equation into the bottom equation and the certain variables cancel out)

"equating to each other" method (reduce both equaations until they are the of same form such as x= [something]... Set the something part to equal the other something part) In the end it is a form of placement, though, I suppose.

"matrices" method. I once knew how to do it but I have long since forgotten how it is supposed to work.
 
  • #7
late347 said:
I see now, but yet I wonder whether or not that is really so much easier
You did not post all your working, but it looked like you would get a quartic equation that would take some work to simplify. My method avoids that, never being worse than a quadratic.
late347 said:
theres about 4 main ways to solve systems of equations
Those methods, as general methods, are for linear equations; these are not.
The equating method does work here, and is what you used. I suppose you could consider my method as an example of the summation method.
 
  • #8
nvm.
 

1. What is a triangle?

A triangle is a polygon with three sides and three angles. It is one of the basic shapes in geometry and is commonly used in various mathematical and scientific fields.

2. How do you find the area of a triangle?

The area of a triangle can be calculated by multiplying the base and height of the triangle and dividing the result by two. The formula for the area of a triangle is: A = 1/2 * base * height.

3. What is the Pythagorean theorem?

The Pythagorean theorem is a fundamental principle in geometry that states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. It can be written as a^2 + b^2 = c^2, where c is the length of the hypotenuse and a and b are the lengths of the other two sides.

4. How do you determine if a triangle is equilateral, isosceles, or scalene?

An equilateral triangle has three sides of equal length, an isosceles triangle has two sides of equal length, and a scalene triangle has no sides of equal length. To determine the type of triangle, you can measure the lengths of all three sides using a ruler or a measuring tape.

5. Can a triangle have more than one right angle?

No, a triangle can only have one right angle (90 degrees). This type of triangle is called a right triangle. If a triangle has three angles that are less than 90 degrees, it is called an acute triangle. If a triangle has one angle that is greater than 90 degrees, it is called an obtuse triangle.

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