Question

show that the point (1,-1,2) is common to the lines which join (6,-7,0) to (16,-19,-4) and(0,3,-6) to (2,-5,10)​

Answer 1

We can see that the point (1, -1, 2) satisfies both the lines. So it is common to both.

Step-by-step explanation:

• First, we write the equation of the line which passes through (6,−7,0) and (16,−19,−4);

• $\dfrac{x-6}{16-6} = \dfrac{y+7}{-19+7} = \dfrac{z-0}{-4-0}$

By further simplification, we get,

• $\dfrac{x-6}{10} = \dfrac{y+7}{-12} = \dfrac{z}{-4}$

Since, point (1, -1, 2) lies on this line. So, this will satisfy the equation of line.

We put x = 1, y = -1, z = 2

• $\dfrac{1-6}{10} = \dfrac{-1+7}{-12} = \dfrac{2}{-4}$

$\dfrac{-1}{2} = \dfrac{-1}{2} = \dfrac{-1}{2}$

• Thus, we observe that the point lies on this line.

In the same way, we write the equation of the line passing through (0, 3, 6) & (16, -19, -4)

• $\dfrac{x-0}{2-0} = \dfrac{y-3}{-5-3} = \dfrac{z+6}{10+6}$

On further simplification, we get.

• $\dfrac{x}{2} = \dfrac{y-3}{-8} = \dfrac{z+6}{16}$

Since, point (1, -1, 2) is common to both line. So, this will satisfy this equation too.

• $\dfrac{1}{2} = \dfrac{-1-3}{-8} = \dfrac{2+6}{16}$

$\dfrac{1}{2} = \dfrac{1}{2} = \dfrac{1}{2}$

Thus, we can see that the point (1, -1, 2) satisfies both the lines. So it is common to both.