Question

write the equation of the plane 4x-4y+2z+5=0 in intercept form​

Answer 1

4x-4y+2z=-5

divide -5 on BS

4x/-5 -4y/-5 +2z/-5=-5/-5

(x/-5/4)+(y/5/4)+(z/-5/2)

Answer 2

The required form $\dfrac{x}{\frac{-5}{4}}+\dfrac{y}{\frac{5}{4}}+\dfrac{z}{\frac{-5}{2}}=1$

Step-by-step explanation:

Given : Equation of plane $4x-4y+2z+5=0$

To find : Write the equation of the plane in intercept form​ ?

Solution :

The intercept form, $\frac{x}{a}+\frac{y}{b} +\frac{z}{c}=1$

Where, a,b,c are the intercept of x,y,z respectively.

Writing the equation as,

$4x-4y+2z=-5$

Divide -5 both side,

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

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

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

#Learn more

Find the equation of the plane which contains the line of intersection of the planes x+2y+3z-4=0 and 2x+y-z+5=0 and whose x intercepts is twice its z intercepts. Hence write the vector equation of the plane passing through the point (2,3,-1) and parallel to the plane obtained earlier

Plz immediately..

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