Question

the area of triangle formed by straight line ax + by + c =0 with coordinate axes is:
​

Answer 1

Answer:

Given equation

ax+by+c=0

On putting y=0 we get x-intercept x=

a

−c

On putting x=0 we get y-intercept y=

b

−c

Area =

a

−c

×

b

−c

=

ab

c

2

Answer by

Sukhmani

Answer 2

Answer:

=> Area of the triangle is C²/2AB

Step-by-step explanation:

The equation of line is Ax +By+C=0,

let the line intersect x- axis ant y-axis at A and B.

At x axis y=0

so Ax=-C, x=-C/A

So the coordinates of A are (-C/A,0)

At y axis x=0

So By=-C, y =(0,-C/B)

Thus ΔOAB , OA=C/A and OB =-CA

also OA ⊥ OB

Thus the area of ΔOAB

=1/2* (C/A)(C/B)=C²/2AB ans.

HOPE IT HELPS..

FOLLOW KROGE MUJHE..

20thanks = Inbox...