Question

What is the equation of the plane that contains the triangle shown in the diagram *answers in question*

Answer 1

Consider, pls, the offered option:
1. knowing points on the graph (2;0;0), (0;5;0) and (0;0;4) it is possible to check an equation.
2. only equation 10x+4y+5z=20 is right if to substitute coordinates in it.
Answer: 10x+4y+5z=20

Answer 2

Answer:

Option C is correct.

$10x+4y+5z =20$

Step-by-step explanation:

The intercept form of the equation of plane is given by:

$\frac{x}{a}+\frac{y}{b} +\frac{z}{c} = 1$    .....[1]

where

a is the intercept of x-axes

b is the intercept of y-axes and

c is the intercept of z-axes

From the given diagram:

a = 2

b = 5

c = 4

Substitute these in equation [1] we have;

$\frac{x}{2}+\frac{y}{5} +\frac{z}{4} = 1$

LCM of 2, 5 and 4 is 20

$\frac{10x+4y+5z}{20} = 1$

Multiply equation 20 both sides we get;

$10x+4y+5z =20$

Therefore, the equation of plane that contains the given triangle is, $10x+4y+5z =20$