Question

Determine graphically, the coordinates of vertices of a triangle whose equations are \(2x - 3y + 6 = 0\); \(2x + 3y - 18 = 0\) and \(x = 0\). Also, find the area of this triangle.

Hint:

When the base of a triangle lies on a coordinate axis, the height is simply the perpendicular distance (absolute coordinate value) of the opposite vertex from that axis.

Answer 1

Step 1: Understanding the Concept:
To solve graphically, we find at least two points for each line and determine their intersection points.
The equation x = 0 represents the y-axis.
The required region is the triangle formed by the two lines and the y-axis.

Step 2: Finding Points for Line 1 (2x − 3y = −6):
If x = 0:
2(0) − 3y = −6
−3y = −6
y = 2
Point = (0, 2)

If y = 0:
2x − 0 = −6
2x = −6
x = −3
Point = (−3, 0)

Step 3: Finding Points for Line 2 (2x + 3y = 18):
If x = 0:
2(0) + 3y = 18
3y = 18
y = 6
Point = (0, 6)

If y = 0:
2x + 0 = 18
2x = 18
x = 9
Point = (9, 0)

Step 4: Finding Intersection of the Two Lines:
Given equations:
2x − 3y = −6
2x + 3y = 18

Add both equations:
4x = 12
x = 3

Substitute in second equation:
2(3) + 3y = 18
6 + 3y = 18
3y = 12
y = 4

Intersection point = (3, 4)

Step 5: Identifying Vertices of the Triangle:
(0, 2)
(0, 6)
(3, 4)

Step 6: Calculating Area:
Base lies on y-axis from y = 2 to y = 6
Base = 6 − 2 = 4 units

Height is horizontal distance from (3, 4) to y-axis
Height = 3 units

Area = 1/2 × Base × Height
= 1/2 × 4 × 3
= 6 square units

Final Answer:
Vertices are (0, 2), (0, 6), (3, 4)
Area of triangle = 6 square units