Question

Three lines x+3y=6 , 2x-3y =12 and x=0 are making triangle . Find the points of intersection of the lines graphically and the area of the park measurement in km

Answer 1

Answer:

18 square km

Step-by-step explanation:

First we find point of intersection of the 3 lines

x+3y=6....(1)

2x-3y =12....(2)

x=0....(3)

From(1)&(2)

3x=18

x=6, y=0

Point of intersection of(1)&(2) is (6,0)

From(2)&(3)

-3y=12

y=-4

Point of intersection of (2)&(3) is(0,-4)

From(3)&(1)

3y=6

y=2

Point of intersection of (3)& (1) is(0,2)

From the diagram,

Area of triangle

=(1/2)(base)(height)

=(1/2)(BC)(OA)

=(1/2)(6)(6)

=18 square km.

Answer 2

Step-by-step explanation:

We have Three lines As : x + 3y = 6 , 2x - 3y = 12 And x = 0

First we form a table for different values of x and y for all three equations As :

x + 3y = 6 2x - 3y = 12 x = 0

x y x y

0 2 0 -4 y -axis

-3 3 -3 -6

-6 4 -6 -8

3 1 3 -2

6 0 6 0

Now we plot these points on x - y plane , As :

Here We have triangular park ABC And point of intersection we get graphically AS :

A ( 6 , 0 ) , B ( 0 , 2 ) And C ( 0 , - 4 )

And Area of triangle we know = 12 ( Base ) ( Height )

So , For Triangle ABC Base BC = 6 km And Height ( y axis to point A ) = 6 km

SO,

Area of triangular park ABC = 12 ( 6 ) ( 6 ) = ( 3 )( 6 ) = 18 km2 ( Ans )