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 Δ OAB (in sq. units) is

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 Δ OAB (in sq. units) is (A) 30 (B) 120 (C) 60 (D) 65

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/bccfdfc953416bb53cd9966c97fc46de-.png" /> Required distance O P=|(0+0+25/√32+42)|=|(25/5)|=5 So, A P=P B=12[ By Pythagoras theorem in Δ A O P] Area of Δ O A B =(1/2) × 24 × 5 =12 × 5=60

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

30

120

60

65

Required distance $OP = ∣ ∣ \frac{0 + 0 + 25}{3 ^{2} + 4 ^{2}} ∣ ∣ = ∣ ∣ \frac{25}{5} ∣ ∣ = 5$
So, $A P = PB = 12 [$ By Pythagoras theorem in $Δ A OP]$
Area of $Δ O A B = \frac{1}{2} \times 24 \times 5$
$= 12 \times 5 = 60$

Q. 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 ΔOAB (in sq. units) is

30

120

60

65

Required distance OP=∣∣​32+42​0+0+25​∣∣​=∣∣​525​∣∣​=5 So, AP=PB=12[ By Pythagoras theorem in ΔAOP] Area of ΔOAB=21​×24×5 =12×5=60

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

Required distance $OP = ∣ ∣ \frac{0 + 0 + 25}{3 ^{2} + 4 ^{2}} ∣ ∣ = ∣ ∣ \frac{25}{5} ∣ ∣ = 5$
So, $A P = PB = 12 [$ By Pythagoras theorem in $Δ A OP]$
Area of $Δ O A B = \frac{1}{2} \times 24 \times 5$
$= 12 \times 5 = 60$