A straight line l1 with equation x-2y+10=0 meets the circle with equation x2+y2=100 at B in the first quadrant. A line through B, perpendicular to l1 cuts the x -axis and y -axis at P and Q respectively. The area (in sq. units) of the triangle OPQ is (where, O is the origin)

Question

A straight line l1 with equation x-2y+10=0 meets the circle with equation x2+y2=100 at B in the first quadrant. A line through B, perpendicular to l1 cuts the x -axis and y -axis at P and Q respectively. The area (in sq. units) of the triangle OPQ is (where, O is the origin) (A) 120 (B) 150 (C) 100 (D) 125

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-wxt0s3waxv0u.png" /> Slope of l1=(1/2) Slope of l2=-2 Equation of l2 is y=-2(x - 10) ⇒ y+2x=20 Hence, P and Q are (10 , 0) and (0 , 20) Area of Δ OPQ=(1/2)× 10× 20=100 sq. units

$120$

$150$

$100$

$125$

Slope of $l_{1} = \frac{1}{2}$
Slope of $l_{2} = - 2$
Equation of $l_{2}$ is $y = - 2 (x - 10)$
$\Rightarrow y + 2 x = 20$
Hence, $P$ and $Q$ are $(10, 0)$ and $(0, 20)$
Area of $Δ OPQ = \frac{1}{2} \times 10 \times 20 = 100$ sq. units

Q. A straight line l1​ with equation x−2y+10=0 meets the circle with equation x2+y2=100 at B in the first quadrant. A line through B, perpendicular to l1​ cuts the x -axis and y -axis at P and Q respectively. The area (in sq. units) of the triangle OPQ is (where, O is the origin)

120

150

100

125

Slope of l1​=21​ Slope of l2​=−2 Equation of l2​ is y=−2(x−10) ⇒y+2x=20 Hence, P and Q are (10,0) and (0,20) Area of ΔOPQ=21​×10×20=100 sq. units

$120$

$150$

$100$

$125$

Slope of $l_{1} = \frac{1}{2}$
Slope of $l_{2} = - 2$
Equation of $l_{2}$ is $y = - 2 (x - 10)$
$\Rightarrow y + 2 x = 20$
Hence, $P$ and $Q$ are $(10, 0)$ and $(0, 20)$
Area of $Δ OPQ = \frac{1}{2} \times 10 \times 20 = 100$ sq. units