Question

1.What is the y-intercept of the plane whose equation is 4x + 5y – z = 20?


(5, 0, 0)
(0, 5, 0)
(0, 4, 0)
(0, 0, 4)

2.Which equation has intercepts at X(1, 0, 0), Y(0, –1, 0), and Z(0, 0, 2)?

3x – 2y + 6z = 12


4x – 4y + z = 4


x – y + 2z = 4


2x – 2y + z = 2


5.is attached below

Answer 1

In case of plane, for y intercept, x and z are o .

So we will get

4(0)+5y-0=20

5y=20

Dividing both sides by 5

y = 4

So the y intercept is (0,4,0).

In second question, x intercept is 1, y intercept is -1 and z intercept is 2 .

So the correct option is the third equation , that is x-y+2z=4 .

Answer 2

Answer:

Option c and

Option d

Step-by-step explanation:

Given is a plane $4x+5y-z=20$

To find y intercept

In the y axis x=0=z

Substitute to get$5y =20:\\y=4$

Hence point is (0,4,0)

2) When x intercept = 1, y int =-1 and z int =2 we have in intercept form the plane

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

is the right one

Option d