Question

A straight line with an equation of 20x + 3y = 12 cuts the x-axis at A and the y-axis at B. Find the coordinates of A and B.

Answer 1

2x+3y=12

6

x

+

4

y

=1

Using point form,

A≡(6,0);B=(0,4)

also, slope of line AB=

3

−2

So, slope of perpendicular =

m

−1

=

2

+3

So, equation will be,

y−5=

2

3

(x−5)

3x−2y=5

5

3x

−

5

2y

=1

Using point form, C≡(

3

5

,0);D≡(0,

2

−5

)

and solving with 2x+3y=12, we get E≡(3,2).

So, O≡(0,0)

C≡(

3

5

,0)

E≡(3,2)

B≡(0,4)

Ar(OCEB)=∣Ar(OCE)∣+∣Ar(OEB)∣

Ar(OCEB)=

2

1

×

3

5

+

2

1

×12=

6

41