Question

If O is the origin and A and B are points
on the line 3x – 4y + 25 = 0 such that
OA = OB = 13 then the area of
AOAB(in sq.units) is​

Answer 1

Answer:

I think it will help you

Answer 2

Answer:

60 square units of the required area of triangle OAB .

Step-by-step explanation:

Explanation:

Given , a line AB  = 3x - 4y + 25 = 0

and OB = OA =  13  and O be the origin (0,0).

Perpendicular distance =   $|\frac{ax+ by + c}{\sqrt{a^2 + b^2} } |$

Let , OP be the perpendicular on AB

Step 1:

We have , 3x - 4y + 25 = 0  .

Therefore , a = 3 , b = -4 and c = 25 .

OP = $|\frac{3 (0) + (-4) (0) + 25 }{\sqrt{3^2 + (-4)^2} } |$ = $|\frac{25}{ \sqrt{9 + 16}} | = \frac{25}{5}$

⇒ OP = 5 units

Now , in right angle triangle  OPA

OP= 5 units , OB = 13 units

By Pythagoras theorem we have ,

$OA^2 = OP^2 + AP ^2$

⇒$AP^2 = OA^2 - OP^2$

⇒AP = $\sqrt{(13)^2 - (5)^2}$ = $\sqrt{ 169 - 25 }$ = $\sqrt{ 144}$

⇒AP = 12 units  

∴ AB = 12 + 12 = 24 units

Step 2:

Area of triangle  = $\frac{1}{2}$ × base × height

⇒Area of ΔOAB = $\frac{1}{2 } (AB ) (OP)$  

[AB is base and OP is height of triangle OAB]

⇒ Area of ΔOAB =  $\frac{1}{2 } ( 24 ) (5)$ = 12 × 5 = 60  square units .

Final answer:

Hence  ,area of triangle OAB is 60 square units .

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