Solution:
Given that the point P lies 1/3 along the segment RS as shown below:
To find the y coordinate of the point P, since the point P lies on 1/3 along the segment RS, we have
Using the section formula expressed as
![[\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bmx_2%2Bnx_1%7D%7Bm%2Bn%7D%2C%5Cfrac%7Bmy_2%2Bny_1%7D%7Bm%2Bn%7D%5D)
In this case,
where
Thus, by substitution, we have
![\begin{gathered} [\frac{1(2)+2(-7)}{1+2},\frac{1(4)+2(-2)}{1+2}] \\ \Rightarrow[\frac{2-14}{3},\frac{4-4}{3}] \\ =[-4,\text{ 0\rbrack} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5B%5Cfrac%7B1%282%29%2B2%28-7%29%7D%7B1%2B2%7D%2C%5Cfrac%7B1%284%29%2B2%28-2%29%7D%7B1%2B2%7D%5D%20%5C%5C%20%5CRightarrow%5B%5Cfrac%7B2-14%7D%7B3%7D%2C%5Cfrac%7B4-4%7D%7B3%7D%5D%20%5C%5C%20%3D%5B-4%2C%5Ctext%7B%200%5Crbrack%7D%20%5Cend%7Bgathered%7D)
Hence, the y-coordinate of the point P is
Answer:
2 real solutions
Step-by-step explanation:
We can use the determinant, which says that for a quadratic of the form ax² + bx + c, we can determine what kind of solutions it has by looking at the determinant of the form:
b² - 4ac
If b² - 4ac > 0, then there are 2 real solutions. If b² - 4ac = 0, then there is 1 real solution. If b² - 4ac < 0, then there are 2 imaginary solutions.
Here, a = 6, b = -20, and c = 1. So, plug these into the determinant formula:
b² - 4ac
(-20)² - 4 * 6 * 1 = 400 - 24 = 376
Since 376 is clearly greater than 0, we know this quadratic has 2 real solutions.
<em>~ an aesthetics lover</em>
1/5 or 5/2 either one works!!
352.7916667 or in another way 352.79
