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Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

A straight li...

Question

A straight line passing through the point A(–2, –3) cuts the line x + 3y = 9 and x + y + 1 = 0 at B and C respectively. If AB.AC = 20, then equation of line can be

x – y = 1

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x – y + 1 = 0

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3x – y + 3 = 0

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3x – y = 3

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Solution

The correct options are
A
x – y = 1

C
3x – y + 3 = 0

Equation of line through A is

y+3=m(x+2) ....(1)

Let point B is $(x_{1}, y_{1})$

$∴ x_{1} = - 2 + A B c o s θ, y_{1} = - 3 + A B s i n θ$

Put $(x_{1}, y_{1})$ in x+3y=9

We get $A B = \frac{20}{c o s θ + 3 s i n θ}$

similarly find $A C = \frac{4}{c o s θ + s i n θ}$

and use AB.AC=20

We get $t a n θ = 1, 3$ put in (1)

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A straight line passing through the point A(–2, –3) cuts the line x + 3y = 9 and x + y + 1 = 0 at B and C respectively. If AB.AC = 20, then equation of line can be

The correct options are A x – y = 1 C 3x – y + 3 = 0 Equation of line through A is y+3=m(x+2) ....(1) Let point B is (x1, y1) ∴x1=−2+AB cosθ, y1=−3+AB sin θ Put (x1, y1) in x+3y=9 We get AB=20cos θ+3 sin θ similarly find AC=4cos θ+sin θ and use AB.AC=20 We get tanθ=1,3 put in (1)