Question

the area of a triangle formed by the lines 2x+y=6,2x-y+2=0 and the x-axis is
a)12sq.units
b)15sq.units
c)10sq.units
d)8sq.units
(class10cbse)​

Answer 1

Answer:

The area of the triangle formed by the lines 2x + y = 6, 2x – y + 2 = 0 and the x – axis is. A: 15sq.

Step-by-step explanation:

Solving the three equations 2x+y=6 2x-y+2=0 and x=0 (equation of x-axis ) we get the three vertices of the triangle to be (1,4); (3,0)and (-1,0) therefore area of the triangle is =1/2 [1 (0-0)+{-1 (0-4)}+3 (4-0)] =1/2×16 =8 (b)

Answer 2

Step-by-step explanation:

Given:

$2x+y=6,2x-y+2 = 0$

To find: The area of a triangle formed by the given lines and the x-axis is

a)12sq.units

b)15sq.units

c)10sq.units

d)8sq.units

Solution:

Tip: Find the intersection of lines and intersection of lines with x-axis.

$\bf Ar(∆ABC) = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Step 1: Find intersection of lines.

$2x + y = 6 \\ 2x - y = - 2 \\ - - - - - - \\ 4x = 4 \\ x = 1$

put value of x in any equation

$2(1) + y = 6 \\ \\ y = 6 - 2 \\ \\ y = 4$

Let the intersection of lines is C(1,4)

Step 2: Find intersection of lines with x-axis.

as we know that y coordinate is zero everywhere on x-axis.

Put y=0

$2x + 0 = 6 \\ \\ x = \frac{6}{2} \\ \\ x = 3$

Let the point is B(3,0)

$2x - (0) = - 2 \\ \\ x = - 1$

let the point is A(-1,0).

Apply the formula to find Area of ∆ABC when three vertices are known.

$Ar(∆ABC) = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

A(-1,0) B(3,0) and C(1,4)

$= \frac{1}{2} | - 1(0 - 4) + 3(4 - 0) + 1(0 - 0)|$

$= \frac{1}{2} |4 + 12| \\ \\ = \frac{1}{2} |16|$

$Ar(∆ABC) = 8 \: sq - units$

Final answer:

Area of triangle is 8 sq-units.

Option D is correct.

To learn more on brainly:

Find the area of the triangle whose vertices are (1,0),(6,0)and(4,3)

https://brainly.in/question/16827510

Draw the graph of the equations x/4+y/5=1 also find the area of the triangle formed by the line and the coordinate axes

...

https://brainly.in/question/46429107