# How to find an equation of the plane through the point (4, -4, -5) and parallel to the plane -3x - 4y + 3z = -5?

**Solution:**

Given point (4, -4, -5)

Equation of the plane passing through the point (4,-4,-5) can be given as

A(x - 4) + B(y + 4) + C(z + 5) = 0 --- (1)

Since the plane is parallel to the plane −3x - 4y + 3z = -5, the coefficients of variables will be in same ratio

So, A/-3 = B/-4 = C/3 = k (K is any constant)

A = -3k, B = -4k, C = 3k

putting the values of A,B,C in eq(1)

-3k(x - 4) - 4k(y + 4) + 3k(z + 5) = 0

-3kx + 12k - 4ky - 16k + 3kz + 15k = 0

-3x + 12 - 4y - 16 + 3z + 15 = 0 [dividing by ‘k’ on both sides]

-3x - 4y + 3z + 11 = 0

3x + 4y - 3z = 11

The equation of plane is 3x + 4y - 3z = 11

## How to find an equation of the plane through the point (4, -4, -5) and parallel to the plane -3x - 4y + 3z = -5?

**Summary:**

The equation of the plane passing through the point (4, -4, -5) and parallel to the plane -3x - 4y + 3z = -5 is 3x + 4y - 3z = 11

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