Question

If the point A and B lying on the straight line 4y =3x + 15 are at a distance 5 units from the origin, find the area of triangle AOB. ​

Answer 1

Answer:

Step-by-step explanation:

first coordinates are -5,0

second coordinates

be x and y

x^2+y^2=25

3x-4y=-15

3x=4y-15

x=4y-15/3

x^2+y^2=25

(4y-15/3)^2 + y^2 = 25

16y^2+225-120y+3y^2=75

19y^2-120y-150=0

y=120 plus minus root 14400+11400) / 38

y=120 plus minus 160.62 / 38

y=-1.06 and  7.38

3x-4y=-15

x=4y-15/3

x=-6.41

x=4.84

therefore points satisfying x^2+y^2=25 and 3x-4y=-15

are (-6.41,-1.06) and (4.84,7.38)

therefore area formed by (-5,0) , (-6.41,-1.06) and (0,0) is

5.3/2

2.65

therefore area formed by (-5,0) and (4.84,7.38) and 0,0 is

36.9/2

18.45

therefore area formed by (4.84,7.38) , (-6.41,-1.06) and 0,0 is

42.17/2

21.085

therefore areas can be three that are

2.65,18.45 and 21.085