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Standard XII

Mathematics

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A straight li...

Question

A straight line $L_{1} : x - 2 y + 10 = 0$ meets the circle with equation $x^{2} + y^{2} = 100$ at $B$ in the first quadrant. If another line $L_{2}$ through $B$ is perpendicular to $L_{1}$ and cuts $y -$ axis at $P (0, t),$ then the value of $t$ is

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Solution

Line $L_{1} : x - 2 y + 10 = 0$
Perpendicular line to $L_{1}$ is $2 x + y + λ = 0$
Now, as $∠ B$ is $90^{\circ},$ so $L_{2}$ must pass through $(10, 0)$
$20 + 0 + λ = 0 \Rightarrow λ = - 20$
Now, $2 x + y - 20 = 0$
$\Rightarrow 0 + t - 20 = 0 ∴ t = 20$

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Standard XII Mathematics

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A straight line L1:x−2y+10=0 meets the circle with equation x2+y2=100 at B in the first quadrant. If another line L2 through B is perpendicular to L1 and cuts y−axis at P(0,t), then the value of t is

Line L1:x−2y+10=0 Perpendicular line to L1 is 2x+y+λ=0 Now, as ∠B is 90∘, so L2 must pass through (10,0) 20+0+λ=0⇒λ=−20 Now, 2x+y−20=0 ⇒0+t−20=0∴t=20

A straight line $L_{1} : x - 2 y + 10 = 0$ meets the circle with equation $x^{2} + y^{2} = 100$ at $B$ in the first quadrant. If another line $L_{2}$ through $B$ is perpendicular to $L_{1}$ and cuts $y -$ axis at $P (0, t),$ then the value of $t$ is

Line $L_{1} : x - 2 y + 10 = 0$
Perpendicular line to $L_{1}$ is $2 x + y + λ = 0$
Now, as $∠ B$ is $90^{\circ},$ so $L_{2}$ must pass through $(10, 0)$
$20 + 0 + λ = 0 \Rightarrow λ = - 20$
Now, $2 x + y - 20 = 0$
$\Rightarrow 0 + t - 20 = 0 ∴ t = 20$