Question

The equation of the line passing through the point of intersection of the lines \(x+2y+6 = 0\) and \(2x-y = 2\) and making an intercept 5 on the y-axis is

• A. \(39x+2y-10 = 0\)
• B. \(39x-2y+10 = 0\)
• C. \(x+4y-20 = 0\)
• D. \(x-4y+20 = 0\)

Hint:

Find the intersection point, then use the two-point form with (0, 5).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A line making y intercept 5 passes through \((0, 5)\). It also passes through the intersection of the two given lines.

Step 2: Find the intersection
From \(2x - y = 2\) we get \(y = 2x - 2\). Put into \(x + 2y + 6 = 0\):
\[ x + 4x - 4 + 6 = 0 \Rightarrow x = -\frac{2}{5},\quad y = -\frac{14}{5} \]

Step 3: Two-point form
Slope between \((0, 5)\) and \(\left(-\frac25, -\frac{14}{5}\right)\):
\[ m = \frac{5 + \frac{14}{5}}{0 + \frac25} = \frac{39/5}{2/5} = \frac{39}{2} \]
Line: \(y = \frac{39}{2}x + 5\), i.e. \(2y = 39x + 10\), so \(39x - 2y + 10 = 0\). All four options have y intercept 5, but only (B) passes through \(\left(-\frac25, -\frac{14}{5}\right)\): \(39(-\frac25) - 2(-\frac{14}{5}) + 10 = -\frac{78}{5} + \frac{28}{5} + 10 = 0\).

Final Answer:
The line is \(39x - 2y + 10 = 0\), option (B). \[ \boxed{39x - 2y + 10 = 0} \]