Question

The equation of a line passing through the point of the intersection of the lines \(x-y+1 = 0\) and \(2x+3y-8 = 0\) and having x intercept \(3\) is

• A. \(x+y+3 = 0\)
• B. \(x+y-3 = 0\)
• C. \(x-y+3 = 0\)
• D. \(-x+y-2 = 0\)

Hint:

Find the intersection point, then use the points $(1,2)$ and $(3,0)$.

Answer 1

The Correct Option is B

Step 1: Find the intersection
From \(x-y+1=0\), \(x=y-1\). Put into \(2x+3y-8=0\): \(2y-2+3y=8\), so \(y=2\) and \(x=1\). The point is \((1,2)\).

Step 2: Use the intercept
x-intercept \(3\) means the line passes through \((3,0)\).

Step 3: Equation
Slope \(=\frac{0-2}{3-1}=-1\). Line: \(y-0=-1(x-3)\), so \(x+y-3=0\).

Step 4: Check
Option (B) is \(x+y-3=0\). Test: \((1,2)\) gives \(1+2-3=0\), and \((3,0)\) gives \(0\). Both lie on it.

Final Answer:
The line is \(x+y-3=0\), option (B). \[ \boxed{\text{(B)}} \]