Question

Solve the equations graphically. 2x+y=2 , 2y-x=4. Also find the area of a triangle formed by the two lines and the line y=0.

Answer 1

Topic :-

Coordinate Geometry

Given :-

2x + y = 2

2y - x = 4

To Find :-

Solution of the equation graphically.

Area of triangle formed by the two lines and the line y = 0.

Concept Used :-

Intersection point of the lines on graph will be the solution for the equations.

$\mathtt{Area\:of\:Triangle=\dfrac{1}{2} \times Base \times Height}$

Solution :-

Drawing line on the graph,

Equation : 2x + y = 2

• Put x = 0 in the equation.

       2(0) + y = 2

       y = 2

• Now, mark point (0, 2) on graph.

• Put y = 0 in the equation.

       2x + 0 = 2

       2x = 2

       x = 1

• Now, mark point (1, 0) on graph.
• Join these two points.

Equation : 2y - x = 4

• Put x = 0 in the equation.

       2y - 0 = 4

       2y = 4

       y = 2

• Now, mark point (0, 2) on graph.

• Put y = 0 in the equation.

       2(0) - x = 4

       -x = 4

       x = -4

• Now, mark point (-4, 0) on graph.
• Join these two points.

Searching point of intersection of lines on graph,

We can observe that (0, 2) is the intersection point of given two lines. Thus, solution of the equations will be x = 0 and y = 2.

Calculating area of triangle formed by two lines and y = 0,

We can observe that a triangle is formed with points (0, 2), (-4, 0) and (1, 0).

Height of the triangle = 2 units

Base of the triangle = 1 - (-4) = 1 + 4 = 5 units

Applying formula for the area of triangle,

$\mathtt{Area\:of\:Triangle=\dfrac{1}{2} \times Base \times Height}$

$\mathtt{Area\:of\:Triangle=\dfrac{1}{\not{2}} \times 5 \times \not{2}\:\:sq.\:units}$

$\mathtt{Area\:of\:Triangle=5 \:\:sq.\:units}$

Answer :-

Solution of the equation is x = 0 and y = 2.

Area of triangle formed is 5 sq. units.

Answer 2

L1 : 2X + Y = 2, L2 : — X + 2Y = 4

Y intercept of both the lines is common (0,2)

X intercepts are (1,0) and (—4,0)

Vertices of triangle formed by intersection of all the three lines are A(—4,0), B(0,2) and C(1,0).

A and C lie on X axis, distance between them is (1-(—4)) 5 units. B lies on Y axis, which is perpendicular to X axis, 2 units is the shortest distance between B and X axis, which is height of the triangle.

So, the area of ∆ABC = 1/2(5)(2) = 5 units^2.