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

Option C is correct.

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

Explanation:

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

`(x)/(a)+(y)/(b) +(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;

`(x)/(2)+(y)/(5) +(z)/(4) = 1`

LCM of 2, 5 and 4 is 20

`(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`