Q32 of 54 Page 77

Solve graphically the pair of linear equations 3x – 4y + 3 = 0 and 3x + 4y – 21 = 0. Find the coordinates of the vertices of the triangular region formed by these lines at X-axis. Also, calculate the area of the triangle.

3x – 4y + 3 = 0

Let x = 1


(3×1) – 4y + 3 = 0


3 + 3 = 4y


6 = 4y



y = 1.5


Let x = 3


3x – 4y + 3 = 0
3×3 – 4y + 3 = 0
9 – 4y = – 3
9 + 3 = 4y


y = 3


Let y = 0
 3x – 4 × 0 = – 3
3x = – 3


x = – 1



3x + 4y – 21 = 0


Let x = 3


3x + 4y – 21 = 0
3×3 + 4y = 21
9 + 4y = 21
4y = 21 – 9
4y = 12



y = 3


Let y = 0


3x + 4×0 = 21
3x + 0 = 21
3x = 21

x = 7


Let x = 1


3×1 + 4y – 21 = 0


3 + 4y – 21 = 0


4y = 18


y = 4.5




Coordinates of ∆ CBE = C (3, 3) B ( – 1, 0) E (7, 0)




Therefore, area of the triangle is 12 square units.


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