Reduce the equations of the following planes in the intercept form and find its intercepts on the coordinate axes: 2x + 3y – z = 6

Asked by Aaryan | 1 year ago |  156

##### Solution :-

Let us reduce the given equation 2x + 3y – z = 6 in intercept form:

Divide the given equation by 6, we get

$$\dfrac{x}{3}+ \dfrac{y}{2}+ \dfrac{z}{-6}=1$$

Now, compare equation 1 and 2, we get

a = 3, b = 2, c = -6

Therefore intercepts on the coordinate axes are 3, 2, -6.

Answered by Sakshi | 1 year ago

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