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Q.

Write the vertices of the triangle formed by the lines x - 2y=4 and $3 x + 5 y = 1$ and the y -axis. Also find the area of this triangle.

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a

Vertices of triangle are (0, 0.2), (0, -2), (2, -1) and Area of triangle

= 2.2 s q

units

b

Vertices of triangle are (0, 2), (0, -2), (2, -1) and Area of triangle

= 5.2 s q

units

c

Vertices of triangle are (0, 1), (0, -2), (2, -1) and Area of triangle

= 2.2 s q

units

d

Vertices of triangle are (0, 3), (0, -2), (2, -1) and Area of triangle

= 3.2 s q

units

answer is A.

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Detailed Solution

Given lines are x - 2y=4 and

3 x + 5 y = 1

.
We know that linear equation in two variables can be solved by graphical method by putting different values of one variable and finding the value of its corresponding variable.
In the first equation, try multiple values of 'x' to get different values of 'y.'
x - 2y = 4
Put, x= 0.

= > − 2 y = 4

⇒ y = − 2

Put, x = 2.

= > 2 − 2 y = 4

⇒

y = -1

x	0	2
y	-2	-1

In the second equation, try multiple values of 'x' to get different values of 'y.'
3x + 5y = 1
Put, x = 0.

= > 5 y = 1

⇒ y = 15

Put, x = 2.

⇒ 32 + 5 y = 1

⇒

y =1

x	0	2
y	$\frac{1}{5}$	-1

Plot the graph of the above two equations with the obtained x and y values.
Question Image

Question Image

As we can observe, a triangle is formed by the given lines and y-axis.
The coordinates of triangle formed are (0, 0.2), (0, -2), (2, -1).

Area of triangle= 12 × Basc × Height

Area of triangle = 12 × 2.2 × 2

= 2.2 sq units
Coordinates of triangle are (0,0.2), (0, -2), (2, -1) and Area of triangle

= 2.2 s q

units.
The correct option is (1).

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