Question

The point at which the line $\frac{x-2}{1}=\frac{y-4}{-5}=\frac{z+3}{4}$ intersects the xy-plane is

• A. $(\frac{11}{4},\frac{1}{4},0)$
• B. $(\frac{5}{4},\frac{1}{4},0)$
• C. $(\frac{11}{4},\frac{3}{4},0)$
• D. $(\frac{7}{4},\frac{1}{4},0)$
• E. $(\frac{11}{4},\frac{7}{4},0)$

Hint:

Geometry Tip: The term "$xy$-plane" means $z$ is missing, so $z=0$. Similarly, the "$yz$-plane" means $x=0$, and the "$xz$-plane" means $y=0$.

Answer 1

The Correct Option is A

Concept:
The $xy$-plane represents the flat surface where height is zero, meaning its equation is simply $z = 0$. To find where a 3D line intersects this plane, we can express the line's coordinates in parametric form using a variable $t$, set the $z$-coordinate to $0$, solve for $t$, and substitute it back.

Step 1: Set the line equation to a parameter t.
Equate the given line ratios to a scalar parameter $t$: $$\frac{x-2}{1} = \frac{y-4}{-5} = \frac{z+3}{4} = t$$

Step 2: Express coordinates in parametric form.
Solve for $x$, $y$, and $z$ in terms of $t$: $$x = t + 2$$ $$y = -5t + 4$$ $$z = 4t - 3$$ Any general point on this line has coordinates $(t+2, -5t+4, 4t-3)$.

Step 3: Apply the condition for the xy-plane.
For the point to lie on the $xy$-plane, its $z$-coordinate must be exactly $0$: $$4t - 3 = 0$$

Step 4: Solve for t.
Isolate the parameter $t$: $$4t = 3 \implies t = \frac{3}{4}$$

Step 5: Find the corresponding x and y coordinates.
Substitute $t = \frac{3}{4}$ back into the equations for $x$ and $y$: $$x = \frac{3}{4} + 2 = \frac{3}{4} + \frac{8}{4} = \frac{11}{4}$$ $$y = -5\left(\frac{3}{4}\right) + 4 = \frac{-15}{4} + \frac{16}{4} = \frac{1}{4}$$ The point of intersection is $(\frac{11}{4}, \frac{1}{4}, 0)$. Hence the correct answer is (A) $(\frac{11}{4},\frac{1}{4},0)$.