Answer:
Distance of the point from its image = 8.56 units
Step-by-step explanation:
Given,
Co-ordinates of point is (-2, 3,-4)
Let's say
![x_1\ =\ -2](https://tex.z-dn.net/?f=x_1%5C%20%3D%5C%20-2)
![y_1\ =\ 3](https://tex.z-dn.net/?f=y_1%5C%20%3D%5C%203)
![z_1\ =\ -4](https://tex.z-dn.net/?f=z_1%5C%20%3D%5C%20-4)
Distance is measure across the line
![\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%2B2%7D%7B3%7D%5C%20%3D%5C%20%5Cdfrac%7B2y%2B3%7D%7B4%7D%5C%20%3D%5C%20%5Cdfrac%7B3z%2B4%7D%7B5%7D)
So, we can write
![\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k](https://tex.z-dn.net/?f=%5Cdfrac%7Bx-x_1%2B2%7D%7B3%7D%5C%20%3D%5C%20%5Cdfrac%7B2%28y-y_1%29%2B3%7D%7B4%7D%5C%20%3D%5C%20%5Cdfrac%7B3%28z-z_1%29%2B4%7D%7B5%7D%5C%20%3D%5C%20k)
![=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k](https://tex.z-dn.net/?f=%3D%3E%5C%20%5Cdfrac%7Bx-%28-2%29%2B2%7D%7B3%7D%5C%20%3D%5C%20%5Cdfrac%7B2%28y-3%29%2B3%7D%7B4%7D%5C%20%3D%5C%20%5Cdfrac%7B3%28z-%28-4%29%29%2B4%7D%7B5%7D%5C%20%3D%5C%20k)
![=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k](https://tex.z-dn.net/?f=%3D%3E%5C%20%5Cdfrac%7Bx%2B4%7D%7B3%7D%5C%20%3D%5C%20%5Cdfrac%7B2y-3%7D%7B4%7D%5C%20%3D%5C%20%5Cdfrac%7B3z%2B16%7D%7B5%7D%5C%20%3D%5C%20k)
![=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}](https://tex.z-dn.net/?f=%3D%3E%5C%20x%5C%20%3D%5C%203k-4%2C%5C%20y%5C%20%3D%5C%20%5Cdfrac%7B4k%2B3%7D%7B2%7D%2C%5C%20z%5C%20%3D%5C%20%5Cdfrac%7B5k-16%7D%7B3%7D)
Since, the equation of plane is given by
x+y+z=3
The point which intersect the point will satisfy the equation of plane.
So, we can write
![3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3](https://tex.z-dn.net/?f=3k-4%2B%5Cdfrac%7B4k%2B3%7D%7B2%7D%2B%5Cdfrac%7B5k-16%7D%7B3%7D%5C%20%3D%5C%203)
![=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18](https://tex.z-dn.net/?f=%3D%3E6%283k-4%29%2B3%284k%2B3%29%2B2%285k-16%29%5C%20%3D%5C%2018)
![=>18k-24+12k+9+10k-32\ =\ 18](https://tex.z-dn.net/?f=%3D%3E18k-24%2B12k%2B9%2B10k-32%5C%20%3D%5C%2018)
![=>\ k\ =\dfrac{13}{8}](https://tex.z-dn.net/?f=%3D%3E%5C%20k%5C%20%3D%5Cdfrac%7B13%7D%7B8%7D)
So,
![x\ =\ 3k-4](https://tex.z-dn.net/?f=x%5C%20%3D%5C%203k-4)
![=\ 3\times \dfrac{13}{8}-4](https://tex.z-dn.net/?f=%3D%5C%203%5Ctimes%20%5Cdfrac%7B13%7D%7B8%7D-4)
![=\ \dfrac{7}{4}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B7%7D%7B4%7D)
![y\ =\ \dfrac{4k+3}{2}](https://tex.z-dn.net/?f=y%5C%20%3D%5C%20%5Cdfrac%7B4k%2B3%7D%7B2%7D)
![=\ \dfrac{4\times \dfrac{13}{8}+3}{2}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B4%5Ctimes%20%5Cdfrac%7B13%7D%7B8%7D%2B3%7D%7B2%7D)
![=\ \dfrac{19}{4}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B19%7D%7B4%7D)
![z\ =\ \dfrac{5k-16}{3}](https://tex.z-dn.net/?f=z%5C%20%3D%5C%20%5Cdfrac%7B5k-16%7D%7B3%7D)
![=\ \dfrac{5\times \dfrac{13}{8}-16}{3}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B5%5Ctimes%20%5Cdfrac%7B13%7D%7B8%7D-16%7D%7B3%7D)
![=\ \dfrac{-21}{8}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B-21%7D%7B8%7D)
Now, the distance of point from the plane is given by,
![d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}](https://tex.z-dn.net/?f=d%5C%20%3D%5C%20%5Csqrt%7B%28x-x_1%29%5E2%2B%28y-y_1%29%5E2%2B%28z-z_1%29%5E2%7D)
![=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}](https://tex.z-dn.net/?f=%3D%5C%20%5Csqrt%7B%28-2-%5Cdfrac%7B7%7D%7B4%7D%29%5E2%2B%283-%5Cdfrac%7B19%7D%7B4%7D%29%5E2%2B%28-4%2B%5Cdfrac%7B21%7D%7B8%7D%29%5E2%7D)
![=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}](https://tex.z-dn.net/?f=%3D%5C%20%5Csqrt%7B%28%5Cdfrac%7B-15%7D%7B4%7D%29%5E2%2B%28%5Cdfrac%7B-7%7D%7B4%7D%29%5E2%2B%28%5Cdfrac%7B9%7D%7B8%7D%29%5E2%7D)
![=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}](https://tex.z-dn.net/?f=%3D%5C%20%5Csqrt%7B%5Cdfrac%7B225%7D%7B16%7D%2B%5Cdfrac%7B49%7D%7B16%7D%2B%5Cdfrac%7B81%7D%7B64%7D%7D)
![=\ \sqrt{\dfrac{1177}{64}}](https://tex.z-dn.net/?f=%3D%5C%20%5Csqrt%7B%5Cdfrac%7B1177%7D%7B64%7D%7D)
![=\ 4.28](https://tex.z-dn.net/?f=%3D%5C%204.28)
So, the distance of the point from its image can be given by,
D = 2d = 2 x 4.28
= 8.56 unit
So, the distance of a point from it's image is 8.56 units.
Answer:
35
Step-by-step explanation:
Answer:
![(x+5)^{2}=-4(y+3)](https://tex.z-dn.net/?f=%28x%2B5%29%5E%7B2%7D%3D-4%28y%2B3%29)
Step-by-step explanation:
Given:
Focus point = (-5, -4)
Vertex point = (-5, -3)
We need to find the equation for the parabola.
Solution:
Since the x-coordinates of the vertex and focus are the same,
so this is a regular vertical parabola, where the x part is squared. Since the vertex is above the focus, this is a right-side down parabola and p is negative.
The vertex of this parabola is at (h, k) and the focus is at (h, k + p). So, directrix is y = k - p.
Substitute y = -4 and k = -3.
![-4 = -3+p](https://tex.z-dn.net/?f=-4%20%3D%20-3%2Bp)
![p=-4+3](https://tex.z-dn.net/?f=p%3D-4%2B3)
![p=-1](https://tex.z-dn.net/?f=p%3D-1)
So the standard form of the parabola is written as.
![(x-h)^{2}=4p(y-k)](https://tex.z-dn.net/?f=%28x-h%29%5E%7B2%7D%3D4p%28y-k%29)
Substitute vertex (h, k) = (-5, -3) and p = -1 in the above standard form of the parabola.
So the standard form of the parabola is written as.
![(x-(-5))^{2}=4(-1)(y-(-3))](https://tex.z-dn.net/?f=%28x-%28-5%29%29%5E%7B2%7D%3D4%28-1%29%28y-%28-3%29%29)
![(x+5)^{2}=-4(y+3)](https://tex.z-dn.net/?f=%28x%2B5%29%5E%7B2%7D%3D-4%28y%2B3%29)
Therefore, equation for the parabola with focus at (-5,-4) and vertex at (-5,-3)
![(x+5)^{2}=-4(y+3)](https://tex.z-dn.net/?f=%28x%2B5%29%5E%7B2%7D%3D-4%28y%2B3%29)
Answer:
Step-by-step explanation:
52-35=17
17/0.5=34