Q) 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

(Q32 – 30/1/1 – CBSE 2026 Question Paper)

Ans: 

Step 1: We have following equations of the lines
2 x – 3 y + 6 = 0
2 x + 3 y – 18 = 0
x = 0

These three lines form the sides of the triangle.

(i) Coordinates of vertices of a triangle:

Step 2: Since any vertex of the triangle  will be the point of intersection of 2 lines.

∴ Let’s take intersection of x = 0 and 2 x – 3 y + 6 = 0 as vertex A

To solve these equations, we will substitute x = 0 in 2 x – 3 y + 6 = 0

∴  2 (0) – 3 y + 6 = 0

∴  3 y = 6

∴  y = 2

So, coordinates of vertex A is (0, 2).

Step 3: Next, let’s take intersection of x = 0 and 2 x + 3 y – 18 = 0 as vertex B

To solve these equations, we will substitute x = 0 in 2 x + 3 y – 18 = 0

∴  2 (0) + 3 y – 18 = 0

∴  3 y = 18

∴  y = 6

So, coordinates of vertex B is (0, 6).

Step 4: Next, let’s take intersection of 2 x – 3 y + 6 = 0 and 2 x + 3 y – 18 = 0 as vertex C

To solve these equations, we will add both equations:

∴  (2 x –  3 y + 6) + (2 x + 3 y – 18) = 0 + 0

∴  2 x – 3 y + 6 + 2 x + 3 y – 18 = 0

∴  4 x – 12 = 0

∴  x = 3

Step 5: Next, we put x = 3 in equation: 2 x – 3 y + 6 = 0

∴ 2 (3) – 3 y + 6 = 0

∴ 6 – 3 y + 6 = 0

∴ 3 y = 12

∴ y = 4

So, coordinates of vertex C is (3,4).

Therefore, the coordinates of the 3 vertices of the triangle are: (0, 2), (0, 6) and (3, 4).

(ii) Area of this triangle:

Let’s plot these 3 points on the graph.

We can see that in the traingle ABC, 2 vertex is on Y-axis and 3rd vertex is in 1st quadrant.

Step 6: Here, Base of triangle = differential of ordinate values of points A and B

= 6 – 2 = 4 units

Step 7: and Height of triangle = abscissa value of point C

= 3 units

Step 8: ∴ Area of triangle = x Base x Height

= x 4 x 3 = 2 x 3

= 6 sq. units

Therefore, the Area of triangle is 6 sq. units.

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