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10 months ago

Straight Lines Class 11th

Let ABC be a triangle with A (-3, 1) and $∠ A C B = θ,$ $0 < θ < \frac{π}{2} .$ If the equation of the median through B is 2x + y – 3 = 0 and the equation of angle bisector of C is 7x – 4y – 1 = 0, then tanq is equal to:

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Vishal Baghel

Contributor-Level 10

Line BM : 2x + y = 3 Þ M (0, 3)

Line CD : 7x – 4y = 1 Þ C (3, 5)

Mirror image of A (-3, 1) in the line CD is $(\frac{13}{5}, \frac{- 11}{5})$ and it will lie on BC.

Slope of AC is 2/3

Slope of BC is 18.

$t a n θ = \frac{18 - \frac{2}{3}}{1 + 18 (\frac{2}{3})} = \frac{4}{3}$

where θ = $∠ A C B$

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New answer posted

10 months ago

Straight Lines Class 11th

A 10 inches long pencil AB with mid point C and a small eraser P are placed on the horizontal top of a table such that PC = $\sqrt[]{5}$ inches and $∠ P C B = t a n^{- 1} (2) .$

The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes exactly 1 inch is:

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alok kumar singh

Contributor-Level 10

$t a n^{- 1} 2 = θ$

->tan q = 2

$s i n (θ - α) = \frac{1}{\sqrt[]{5}}$

->4 – 2 tanq = 1 + 2 tan a tan a = $\frac{3}{4}$

$α = t a n^{- 1} \frac{3}{4}$

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New answer posted

10 months ago

Straight Lines Class 11th

Let A(, 2), B(, 6) and $C (\frac{α}{4}, - 2)$ be vertices of a $Δ A B C .$ If $(5, \frac{α}{4})$ is the circumcentre of $Δ A B C,$ then which of the following is NOT correct about $Δ A B C ?$

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Payal Gupta

Contributor-Level 10

Circumcentre (D) $\equiv (5, \frac{α}{4})$

${(5 - α)}^{2} + {(\frac{α}{4} + 2)}^{2} = {(5 - α)}^{2} + {(\frac{α}{4} - 6)}^{2} . . . . . . . . . . . . . . . (i)$

${(5 - \frac{α}{4})}^{2} + {(\frac{α}{4} + 2)}^{2} . . . . . . . . . . . . . . . . . . . . (i i)$

(i) $\frac{α}{4} + 2 = \pm (\frac{α}{4} - 6)$

(ii) 9 + 16 = 9 + 16

$\oplus \to x$

$(-) \to \frac{α}{2} = 4 \Rightarrow α = 8$

ar $(A B C) = 24$

2S = 24

R = 5, r = $\frac{Δ}{s} = \frac{24}{12} = 2$

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New answer posted

10 months ago

Straight Lines Class 11th

Let m₁, m₂ be the slopes of two adjacent sides of a square of side a such that a²+ 11a + $3 (m_{1}^{2} + m_{2}^{2}) = 220 .$ If one vertex of the square is $(10 (c o s α - s i n α), 10 (s i n α + c o s α)),$ where $α \in (0, \frac{π}{2})$ and the equation of one diagonal is $(c o s α - s i n α) x + (s i n α + c o s α) y = 10,$ then $72 (s i n^{4} α + c o s^{4} α) + a^{2} - 3 a + 13$ is equal to

0 Follower 22 Views

Payal Gupta

Contributor-Level 10

$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cosα - sinα)x + (sinα + cosα)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a²+ 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2α - 10cos2α = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

FIGURE

? = $72 (\frac{1}{16} + \frac{9}{16}) + 100 - 30 + 13 = \frac{720}{16} + 83 = 128$

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New answer posted

10 months ago

Straight Lines Class 11th

A circle touches both the y-axis and the line x + y = 0. Then the locus of its center is:

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alok kumar singh

Contributor-Level 10

$| h | = r a n d \frac{| h | + k_{1}}{\sqrt[]{2}} = r$

\frac{{(h + k)}^{2}}{2} = h^{2} \Rightarrow h^{2} = k^{2} + 2 h k

x^{2} - y^{2} = 2 x y

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New answer posted

10 months ago

Straight Lines Class 11th

Let A(1, 1) B(-4, 3), C(-2, -5) be vertices of a triangle ABC, P be a point on side BC, and $Δ_{1} a n d Δ_{2}$ be the area of triangles APB and ABC, respectively. If $Δ_{1} : Δ_{2} = 4 : 7,$ then the area enclosed by the lines AP, AC and the x-axis is

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alok kumar singh

Contributor-Level 10

$\frac{Δ_{1}}{Δ_{2}} = \frac{| \begin{matrix} 1 & 1 & 1 \\ x & - 4 x - x & 1 \\ - 4 & 3 & 1 \end{matrix} |}{| \begin{matrix} 1 & 1 & 1 \\ - 4 & 3 & 1 \\ - 2 & - 5 & 1 \end{matrix} |} = \frac{4}{7}$

->14 x – 35 y = -95 …. (ii)

Solve (i) & (ii), x = $\frac{- 20}{7}, y = \frac{- 11}{7}$

$a r Δ A Q R$

$= \frac{1}{2} * 1 * 1 = \frac{1}{2}$

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New answer posted

10 months ago

Straight Lines Class 11th

Let

A₁ = ${(x, y) : | x | \leq y^{2}, | x | + 2 y \leq 8} a n d$

A₂ = ${(x, y) : | x | + | y | \leq k} :$ If 27 (Area A₁) = 5 (Area A₂), then k is equal to:

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alok kumar singh

Contributor-Level 10

Required area (above x-axis)

$A_{1} = 2 \int_{0}^{4} (\frac{8 - x}{2} - \sqrt[]{x}) d x$

$= 2 (16 - \frac{16}{4} - \frac{8}{3 / 2}) = \frac{40}{3}$

and $A_{2} = 4 (\frac{1}{2} . k^{2}) = 2 k^{2}$

$∴ 27 . \frac{40}{3} = 5 . (2 k^{2})$

-> k = 6

for above x-axis.

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New answer posted

10 months ago

Straight Lines Class 11th

Let $\frac{d y}{d x} = \frac{a x - b y + a}{b x + c y + a},$ where a, b, c, are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is:

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

$\frac{d y}{d x} = \frac{a x - b y + a}{b x + c y + a}$

$= b x d y + c y d y + a d y = a x d x - b y d x + a d x$

$= \frac{c y^{2}}{2} + a y - \frac{a x^{2}}{2} - a x + b x y = k$

$a x^{2} + a y^{2} + 2 a x - 2 a y = k$

$\Rightarrow x^{2} + y^{2} + 2 x - 2 y = λ$

Short distance of (11,6)

= $\sqrt[]{1 2^{2} + 5^{2}} - 5$

= 13 – 5

= 8

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New answer posted

10 months ago

Straight Lines Class 11th

Let A(a, -2), B(a, 6) and $C (\frac{α}{4}, - 2)$ be vertices of a $Δ A B C .$ If $(5, \frac{α}{4})$ is the circumcentre of $Δ A B C,$ then which of the following is NOT correct about $Δ A B C ?$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

Circumcentre (D) $\equiv (5, \frac{α}{4})$

${(5 - α)}^{2} + {(\frac{α}{4} + 2)}^{2} = {(5 - α)}^{2} + {(\frac{α}{4} - 6)}^{2} . . . . . . . . . . . . . . . (i)$

${(5 - \frac{α}{4})}^{2} + {(\frac{α}{4} + 2)}^{2} . . . . . . . . . . . . . . . . . . . . (i i)$

(i) -> $\frac{α}{4} + 2 = \pm (\frac{α}{4} - 6)$

(ii) -> 9 + 16 = 9 + 16

$\oplus \to x$

$(-) \to \frac{α}{2} = 4 \Rightarrow α = 8$

ar $(A B C) = 24$

2S = 24

R = 5, r = $\frac{Δ}{s} = \frac{24}{12} = 2$

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New answer posted

10 months ago

Straight Lines Class 11th

Let m₁, m₂ be the slopes of two adjacent sides of a square of side a such that a² + 11a + $3 (m_{1}^{2} + m_{2}^{2}) = 220 .$ If one vertex of the square is $(10 (c o s α - s i n α), 10 (s i n α + c o s α)),$ where $α \in (0, \frac{π}{2})$ and the equation of one diagonal is $(c o s α - s i n α) x + (s i n α + c o s α) y = 10,$ then $72 (s i n^{4} α + c o s^{4} α) + a^{2} - 3 a + 13$ is equal to:

0 Follower 14 Views

alok kumar singh

Contributor-Level 10

$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cos a - sina)x + (sina + cosa)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a² + 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

-> 210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

->m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2a - 10cos2a = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

...more

$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cos a - sina)x + (sina + cosa)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a² + 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

-> 210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

->m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2a - 10cos2a = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

FIGURE

? = $72 (\frac{1}{16} + \frac{9}{16}) + 100 - 30 + 13 = \frac{720}{16} + 83 = 128$

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