Question

The point of intersection of the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-11}{4}$ and $\frac{x-3}{1}=\frac{y-\frac{9}{2}}{2}=\frac{z}{1}$ is ________.

• A. $(-2,\frac{11}{2},5)$
• B. $(-2,\frac{11}{2},-5)$
• C. $(-2,\frac{-11}{2},-5)$
• D. $(-2,\frac{11}{4},-5)$
• E. $(-2,\frac{-11}{5},\frac{-5}{2})$

Hint:

To find the intersection of two lines, convert them to parametric form and solve for the parameters $r$ and $k$.

Answer 1

The Correct Option is C

Step 1: Parametric Form
Let $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-11}{4} = r \implies (2r+1, 3r-1, 4r+11)$
Let $L_2: \frac{x-3}{1}=\frac{y-4.5}{2}=\frac{z}{1} = k \implies (k+3, 2k+4.5, k)$

Step 2: Equating x and y
$2r + 1 = k + 3 \implies k = 2r - 2$
$3r - 1 = 2k + 4.5 \implies 3r - 1 = 2(2r - 2) + 4.5 = 4r + 0.5$
$r = -1.5$.

Step 3: Calculating Point
For $r = -1.5$: $x = 2(-1.5)+1 = -2$; $y = 3(-1.5)-1 = -5.5 = -11/2$.
For $k = 2(-1.5)-2 = -5$: $x = -2$; $y = -5.5$; $z = -5$.

Step 4: Conclusion
The point $(-2, -11/2, -5)$ satisfies the equations for $x$ and $y$ on both lines and matches the third coordinate for $L_2$. Final Answer: (C)