Question

Draw the graph of pair of linear equations x-y+2=0 and 4x-y-4=0. calculate the area of triangle formed by lines so drawn and x-axis.

Answer 1

Given: Linear equation x-y+2=0 and 4x-y-4=0.

To find : Area of triangle formed by lines so drawn and x-axis

Step-by-step explanation:

1) x-y+2=0

if x = 0

$0-y+2=0\\\\\Rightarrow y= 2$

so (0,2) is the solution

Now if y=0

$x-0+2=0\\\\\Rightarrow x=-2$

Hence, (-2,0) is the solution

Similarly

4x-y-4=0

(0,-4) and (1,0) are the solution

Attached graph

Now as we can see from the graph

The coordinate of triangle formed by lines so drawn and x-axis are (-2,0),(1,0), (2,4)

Now as we know area of triangle is given by

$Area=\dfrac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

where

$(x_1,y_1)=(-2,0) ,(x_2,y_2)=(1,0), (x_3,y_3)=(2,4)$

So we have

$Area = \dfrac{1}{2} |-2(0-4)+1(4-0)+2(0-0)|\\\\\Rightarrow Area = \dfrac{1}{2} |8+4|=\dfrac{12}{2} =6$

Hence, the area of  required triangle is 6 square units

Answer 2

Step-by-step explanation:

x-y=2

X=0

Y=2

Y=0

X=2

Similarly,

4x-y-4 =0

X=0

Y=-4

Y=0 then

X=1

Hence the coordinates are (1,0) & (0,-4)