The equation of the pair of lines perpendicular to the lines given equation is
.
Solution:
Given equation is
.
Let
and
be the slopes of the given lines.
Sum of the roots = ![-\frac{\text {coefficient of } x y}{\text {coefficient of } y^{2}}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Ctext%20%7Bcoefficient%20of%20%7D%20x%20y%7D%7B%5Ctext%20%7Bcoefficient%20of%20%7D%20y%5E%7B2%7D%7D)
– – – – – (1)
Product of the roots = ![-\frac{\text {coefficient of } x^2}{\text {coefficient of } y^{2}}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Ctext%20%7Bcoefficient%20of%20%7D%20x%5E2%7D%7B%5Ctext%20%7Bcoefficient%20of%20%7D%20y%5E%7B2%7D%7D)
– – – – – (2)
The required lines are perpendicular to these lines.
Slopes of the required lines are ![$-\frac{1}{m_{1}} \text { and }-\frac{1}{m_{2}}](https://tex.z-dn.net/?f=%24-%5Cfrac%7B1%7D%7Bm_%7B1%7D%7D%20%5Ctext%20%7B%20and%20%7D-%5Cfrac%7B1%7D%7Bm_%7B2%7D%7D)
Required lines also passes through the origin,
therefore their y-intercepts are 0.
Hence their equations are:
![$y=-\frac{1}{m_{1}}x \text { and }y=-\frac{1}{m_{2}}x](https://tex.z-dn.net/?f=%24y%3D-%5Cfrac%7B1%7D%7Bm_%7B1%7D%7Dx%20%5Ctext%20%7B%20and%20%7Dy%3D-%5Cfrac%7B1%7D%7Bm_%7B2%7D%7Dx)
Do cross multiplication, we get
![m_1y=-x \ \text{and} \ m_2y=-x](https://tex.z-dn.net/?f=m_1y%3D-x%20%5C%20%20%5Ctext%7Band%7D%20%5C%20%20m_2y%3D-x)
Add x on both sides of the equation, we get
![x+m_1y=0 \ \text{and} \ x+m_2y=0](https://tex.z-dn.net/?f=x%2Bm_1y%3D0%20%5C%20%20%5Ctext%7Band%7D%20%5C%20%20x%2Bm_2y%3D0)
Therefore, the joint equation of the line is
![\left(x+m_{1} y\right)\left(x+m_{2} y\right)=0](https://tex.z-dn.net/?f=%5Cleft%28x%2Bm_%7B1%7D%20y%5Cright%29%5Cleft%28x%2Bm_%7B2%7D%20y%5Cright%29%3D0)
![x^2+m_2xy+m_1xy+m_1m_2y^2=0](https://tex.z-dn.net/?f=x%5E2%2Bm_2xy%2Bm_1xy%2Bm_1m_2y%5E2%3D0)
![x^{2}+\left(m_{1}+m_{2}\right) x y+m_{1} m_{2} y^{2}=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B%5Cleft%28m_%7B1%7D%2Bm_%7B2%7D%5Cright%29%20x%20y%2Bm_%7B1%7D%20m_%7B2%7D%20y%5E%7B2%7D%3D0)
Substitute (1) and (2), we get
![$x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0](https://tex.z-dn.net/?f=%24x%5E%7B2%7D%2B%5Cleft%28%5Cfrac%7B-2%20h%7D%7Bb%7D%5Cright%29%20x%20y%2B%5Cleft%28%5Cfrac%7Ba%7D%7Bb%7D%5Cright%29%20y%5E%7B2%7D%3D0)
To make the denominator same, multiply and divide first term by b.
![$ \frac{b}{b} x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0](https://tex.z-dn.net/?f=%24%20%5Cfrac%7Bb%7D%7Bb%7D%20x%5E%7B2%7D%2B%5Cleft%28%5Cfrac%7B-2%20h%7D%7Bb%7D%5Cright%29%20x%20y%2B%5Cleft%28%5Cfrac%7Ba%7D%7Bb%7D%5Cright%29%20y%5E%7B2%7D%3D0)
![$\frac{bx^2-2hxy+ay^2}{b} = 0](https://tex.z-dn.net/?f=%24%5Cfrac%7Bbx%5E2-2hxy%2Bay%5E2%7D%7Bb%7D%20%3D%200)
Do cross multiplication, we get
![b x^{2}-2 h x y+a y^{2}=0](https://tex.z-dn.net/?f=b%20x%5E%7B2%7D-2%20h%20x%20y%2Ba%20y%5E%7B2%7D%3D0)
Hence equation of the pair of lines perpendicular to the lines given equation is
.
Answer:
In the graph, we can see that the relation between length and weight is given by the adjusted line, which passes through the points (24, 16) and (28, 25)
Remember that a linear relation can be written as:
y = a*x + b
Where a is the slope and b is the y-intercept.
If this line passes through the points (x₁, y₁) and (x₂, y₂) the slope can be calculated as:
a = (y₂ - y₁)/(x₂ - x₁)
Then in our case, the slope will be:
a = (25 - 16)/(28 - 24) = 9/4
y = (9/4)*x + b
Knowing that this line passes through (24, 16), we know that when x = 24, y must be equal to 16.
If we replace these in the equation, we can find the value of b.
16 = (9/4)*24 + b
16 = 54 + b
16 - 54 = b - 38
Then the equation is:
y = (9/4)*x - 38
Now that we know the equation, we can simply replace y by 34 pounds to find the value of x.
34 = (9/4)*x - 38
34 + 38 = (9/4)*x
72*(4/9) = x = 32
So we can estimate that the length of a fish that weighs 34 pounds is 32 (I do not know the unit of length, I can't see the horizontal axis on the image)
From the sizes give,
there are:
3 - 9's
3 - 8's
1 - 7
2 -10's
1-11
the second graph has the right amount of x's for the given shoe sizes
Ok, 7, 6, 5, and 4 are the integers.
Work - So, if you add 7 and 6 you get 13, and the problem tells you to multiply that by 2. After you multiply by 2, you should get 26, after getting that you move 2 numbers before 7 and 6, you should get to 4, the problem also tells us to make sure you multiply it by 3 but make it 14 less than you answer. Therefore, after doing the work you should receive the number 12, if you get your previous number, in our case, 26, you need to then, subtract 12 away from 26, and you should get 14 (the number you should be left with).
This is correct, Have a good day Mr. Abc