Question

if p, q are two points on the line 3x+4y+15=0 such that op=oq=9 then area of triangle opq
is
​

Answer 1

if p, q are two points on the line 3x+4y+15=0 such that op=oq=9 then area of triangle opq

is

​

Step-by-step explanation:

Given in △OPQ, O is the origin and OP=OQ=9

Therefore, △OPQ is an isosceles  triangle.

Let, C be the midpoint of PQ, OC is the perpendicular to the line PQ.

OC= Length of the perpendicular from (0,0) to the line 3x+4y+15=0

⇒OC=  

 ∣0+0+15∣   / √ 3²  + √4 ²

​    15 /√ 25 = 15  / 5  =3

In the right angled triangle OCP,

OP²=OC²+PC  ²

 ⇒9  =3²  +PC²

⇒PC² =81−9

⇒PC²  =72

⇒PC=√  72 =6 √ 2

​    Hence, C is the midpoint of PQ

∴PQ=2×6  √2

​ =12  √2

​  

 ∴ Area of △OPQ=  1/2 ×OC×PQ

=  1 /2×3×12 √2

​  

=18  √2 square units.