The equation of the line joining the points (-2, 4, 2) and (7, -2, 5) are

Question

The equation of the line joining the points (-2, 4, 2) and (7, -2, 5) are (A)  (x /-2) = (y /4) = (z /2) (B)  (x /7) = (y /-2) = (z /5) (C)  (x+2 /3) = (y-4 /-2) = (z-2/1) (D) none of these

Tardigrade Admin · Accepted Answer

Line is (x+2/7+2)=(y-4/2-4)=(z-2/5-2) i.e., (x+2/9)=(y-4/-6)=(z-2/3) i.e., (x+2/3)=(y-4/-2)=(z-2/1)

Q. The equation of the line joining the points (-2, 4, 2) and (7, -2, 5) are

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

$\frac{x}{7} = \frac{y}{- 2} = \frac{z}{5}$

$\frac{x + 2}{3} = \frac{y - 4}{- 2} = \frac{z - 2}{1}$

none of these

Line is $\frac{x + 2}{7 + 2} = \frac{y - 4}{2 - 4} = \frac{z - 2}{5 - 2}$
i.e., $\frac{x + 2}{9} = \frac{y - 4}{- 6} = \frac{z - 2}{3}$
i.e., $\frac{x + 2}{3} = \frac{y - 4}{- 2} = \frac{z - 2}{1}$

Q. The equation of the line joining the points (-2, 4, 2) and (7, -2, 5) are

−2x​=4y​=2z​

7x​=−2y​=5z​

3x+2​=−2y−4​=1z−2​

none of these

Line is 7+2x+2​=2−4y−4​=5−2z−2​ i.e., 9x+2​=−6y−4​=3z−2​ i.e., 3x+2​=−2y−4​=1z−2​

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

$\frac{x}{7} = \frac{y}{- 2} = \frac{z}{5}$

$\frac{x + 2}{3} = \frac{y - 4}{- 2} = \frac{z - 2}{1}$

Line is $\frac{x + 2}{7 + 2} = \frac{y - 4}{2 - 4} = \frac{z - 2}{5 - 2}$
i.e., $\frac{x + 2}{9} = \frac{y - 4}{- 6} = \frac{z - 2}{3}$
i.e., $\frac{x + 2}{3} = \frac{y - 4}{- 2} = \frac{z - 2}{1}$