Question

The lines $3x-2y=-6$ and $x+y=3$ meet the $Y$-axis at the point $A$ and meet the $X$-axis at the points $B$ and $C$. Then the area of $\triangle ABC$ (in sq. units) is

• A. 15
• B. 7.5
• C. 6
• D. 8

Hint:

Whenever two lines have the same Y-intercept, first find that common point and use it as the vertex on the Y-axis. Here: \[ A=(0,3), \quad B=(-2,0), \quad C=(3,0) \] The base is simply \[ BC=3-(-2)=5 \] and the height is \[ 3 \] Thus, \[ \text{Area} = \frac{1}{2} \times 5 \times 3 = 7.5 \] This is much faster than using the coordinate geometry determinant formula.

Answer 1

The Correct Option is B

Concept: To determine the area of a triangle formed by the intercepts of straight lines on the coordinate axes, we first identify the coordinates of the points where the given lines intersect the axes. For any straight line:
• The Y-intercept is obtained by putting $x=0$.
• The X-intercept is obtained by putting $y=0$. Once the coordinates of the vertices are known, the area of the triangle can be calculated using: \[ \text{Area of Triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} \] Since two vertices lie on the X-axis and one vertex lies on the Y-axis, the computation becomes straightforward.

Step 1: Find the point where the lines meet the Y-axis. The first line is \[ 3x-2y=-6 \] To find its Y-intercept, put $x=0$: \[ 3(0)-2y=-6 \] \[ -2y=-6 \] \[ y=3 \] Therefore, this line meets the Y-axis at \[ (0,3) \] Now consider the second line \[ x+y=3 \] Putting $x=0$: \[ 0+y=3 \] \[ y=3 \] Hence, the second line also meets the Y-axis at \[ (0,3) \] Therefore, both lines intersect the Y-axis at the same point. Thus, \[ A=(0,3) \]

Step 2: Find the X-intercept of the first line. For the line \[ 3x-2y=-6 \] put \[ y=0 \] Then \[ 3x=-6 \] \[ x=-2 \] Hence, the point where this line meets the X-axis is \[ B=(-2,0) \]

Step 3: Find the X-intercept of the second line. For the line \[ x+y=3 \] put \[ y=0 \] Then \[ x=3 \] Hence, the point where this line meets the X-axis is \[ C=(3,0) \]

Step 4: Calculate the length of the base $BC$. Since both $B$ and $C$ lie on the X-axis, the length of $BC$ is simply the difference of their X-coordinates. \[ BC = |3-(-2)| \] \[ BC = 5 \] Thus, \[ \text{Base}=5 \text{ units} \]

Step 5: Determine the height of the triangle. The height of the triangle is the perpendicular distance of point $A(0,3)$ from the X-axis. Since the X-axis has equation $y=0$, the perpendicular distance is simply the absolute value of the Y-coordinate of $A$. \[ \text{Height} = |3| = 3 \text{ units} \]

Step 6: Calculate the area of $\triangle ABC$. Using the formula \[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \] Substituting the values obtained above: \[ \text{Area} = \frac{1}{2} \times 5 \times 3 \] \[ = \frac{15}{2} \] \[ = 7.5 \] Therefore, \[ \boxed{\text{Area of } \triangle ABC = 7.5 \text{ square units}} \] Hence, the correct answer is \[ \boxed{\text{(B) } 7.5} \]