Question

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

Answer 1

Answer:

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

Therefore,△OPQ is isosceles

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

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

⇒OC=

3

2

+4

2

∣0+0+15∣

=

25

15

=

5

15

=3

In right angled triangle OCP,

OP

2

=OC

2

+PC

2

⇒9

2

=3

2

+PC

2

⇒PC

2

=81−9

⇒PC

2

=72

⇒PC=

72

=6

2

Since, C is the midpoint of PQ

∴PQ=2×6

2

=12

2

∴ Area of △OPQ=

2

1

×OC×PQ

=

2

1

×3×12

2

=18

2

squareunits.