Question

Show that the points (1, 1, 1) and (-3, 0, 1) are
equidistant from the plane 3x + 4y - 12z + 13 = 0.​

Answer 1

Step-by-step explanation:

As we know that distance of point (x₁ , y₁ , x₁ ) From the plane ax + by + cz + d = 0 is given by

|ax₁ + by₁ + cz₁ + d| / √a² + b² + c² = 0

Distance of the point (1,1,1) from the plane 3x+4y-12z+ 13=0

The required distance

• = |3(1)+ 4(1) + (1)+ 13| / √a² +b² + c²
• = |3+4+-12(1)+13|/√3²+4²+(-12)²
• = |8|/√9+16+144
• 8/13 units _____(1)

Distance of the point (-3, 0, 1) from the plane 3x +4y -12z +13 = 0

The required distance

• |3(-3)+4(0)-12(1)+13|/√3²+4² + (-12)²
• |-9+0-12+13|/√169
• |-8|/13 = 8/13 ______(2)

From (1) and (2) we can say that point (1,1,1) and (-3,0,1) are equidistant from the plane 3x+4y-12z +13 = 0

______________________________

Answer 2

Step-by-step explanation:

here is your answer

Hope it helps you