Question

The sides of a triangle are \( 3x + 2y - 6 = 0 \), \( 2x - 3y + 6 = 0 \) and \( x + 2y + 2 = 0 \). If \( P(0, b) \) lies either on the triangle or inside the triangle, then \( b \) lies in the interval

• A. \( [-1, 3] \)
• B. \( [2, 3] \)
• C. \( [-1, 2] \)
• D. \( [-2, 2] \)

Hint:

Find the vertices of the triangle by solving the pairs of linear equations. For a point to lie inside a triangle, the sign of the equation of each side evaluated at the point should be the same as the sign evaluated at the opposite vertex (or consistently opposite to a reference point outside). Check the signs of \( 3x + 2y - 6 \), \( 2x - 3y + 6 \), and \( x + 2y + 2 \) at the vertices and at \( P(0, b) \) to determine the range of \( b \).

Answer 1

The Correct Option is C

The vertices of the triangle are the intersection points of the three lines.
Intersection of \( 3x + 2y - 6 = 0 \) and \( 2x - 3y + 6 = 0 \): Multiply the first by 3 and the second by 2: \( 9x + 6y - 18 = 0 \) \( 4x - 6y + 12 = 0 \) Adding these gives \( 13x - 6 = 0 \implies x = \frac{6}{13} \).
Substituting \( x = \frac{6}{13} \) into \( 3x + 2y - 6 = 0 \): \( 3(\frac{6}{13}) + 2y - 6 = 0 \implies \frac{18}{13} + 2y - \frac{78}{13} = 0 \implies 2y = \frac{60}{13} \implies y = \frac{30}{13} \).
Vertex A: \( (\frac{6}{13}, \frac{30}{13}) \) Intersection of \( 2x - 3y + 6 = 0 \) and \( x + 2y + 2 = 0 \): From the second equation, \( x = -2y - 2 \).
Substitute into the first: \( 2(-2y - 2) - 3y + 6 = 0 \implies -4y - 4 - 3y + 6 = 0 \implies -7y + 2 = 0 \implies y = \frac{2}{7} \).
\( x = -2(\frac{2}{7}) - 2 = -\frac{4}{7} - \frac{14}{7} = -\frac{18}{7} \).
Vertex B: \( (-\frac{18}{7}, \frac{2}{7}) \) Intersection of \( 3x + 2y - 6 = 0 \) and \( x + 2y + 2 = 0 \): Subtracting the second from the first: \( (3x + 2y - 6) - (x + 2y + 2) = 0 \implies 2x - 8 = 0 \implies x = 4 \).
Substituting \( x = 4 \) into \( x + 2y + 2 = 0 \): \( 4 + 2y + 2 = 0 \implies 2y = -6 \implies y = -3 \).
Vertex C: \( (4, -3) \) The point \( P(0, b) \) lies on or inside the triangle.
We need to check the sign of \( f(x, y) \) for each line at the origin (or any point not on the line) and at \( P(0, b) \).
Line 1: \( 3x + 2y - 6 = 0 \).
At origin: \( -6<0 \).
At \( P(0, b) \): \( 2b - 6 \).
Line 2: \( 2x - 3y + 6 = 0 \).
At origin: \( 6>0 \).
At \( P(0, b) \): \( -3b + 6 \).
Line 3: \( x + 2y + 2 = 0 \).
At origin: \( 2>0 \).
At \( P(0, b) \): \( 2b + 2 \).
Check the signs at the vertices: A \( (\frac{6}{13}, \frac{30}{13}) \): \( \frac{18}{13} + \frac{60}{13} - 6 = 0 \), \( \frac{12}{13} - \frac{90}{13} + 6 = 0 \), \( \frac{6}{13} + \frac{60}{13} + 2 \neq 0 \) (Error in vertex calculation) Let's recheck vertex A: \( 9x + 6y = 18 \) \( 4x - 6y = -12 \) Adding: \( 13x = 6 \implies x = 6/13 \).
\( 2(6/13) - 3y + 6 = 0 \implies 12/13 + 78/13 = 3y \implies 90/13 = 3y \implies y = 30/13 \).
Correct.
We need to ensure \( P(0, b) \) satisfies the inequalities defined by the interior of the triangle.
Consider the signs of the expressions at the vertices.
At A \( (\frac{6}{13}, \frac{30}{13}) \): \( 3x + 2y - 6 = 0 \), \( 2x - 3y + 6 = 0 \), \( x + 2y + 2 = \frac{6}{13} + \frac{60}{13} + 2>0 \) At B \( (-\frac{18}{7}, \frac{2}{7}) \): \( 3x + 2y - 6 = -\frac{54}{7} + \frac{4}{7} - 6<0 \), \( 2x - 3y + 6 = 0 \), \( x + 2y + 2 = -\frac{18}{7} + \frac{4}{7} + 2 = 0 \) At C \( (4, -3) \): \( 3x + 2y - 6 = 12 - 6 - 6 = 0 \), \( 2x - 3y + 6 = 8 + 9 + 6>0 \), \( x + 2y + 2 = 4 - 6 + 2 = 0 \) For \( P(0, b) \) to be inside: Sign of \( 3x + 2y - 6 \) at \( P \) should be the same as at B: \( 2b - 6 \le 0 \implies b \le 3 \) Sign of \( 2x - 3y + 6 \) at \( P \) should be the same as at C: \( -3b + 6 \ge 0 \implies b \le 2 \) Sign of \( x + 2y + 2 \) at \( P \) should be the same as at A: \( 2b + 2 \ge 0 \implies b \ge -1 \) So, \( -1 \le b \le 2 \).