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2 months ago

Straight Lines Class 11th

Let A be the set of all points (α, β) such that the area of triangle formed by the points (5, 6), (3, 2) and (α, β) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is :

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Vishal Baghel

Contributor-Level 10

$| \begin{matrix} α & β & 1 \\ 5 & 6 & 1 \\ 3 & 2 & 1 \end{matrix} | = 24$

$4 α - 2 β - 8 = \pm 24$

$4 α - 2 β = 24 + 8, 4 α - 2 β = - 24 + 8$

$2 (2 α - β) = 32 2 α - β = - 8$

Distance of origin

$D = \sqrt[]{α^{2} + {(2 α + 8)}^{2}}$

$α = - \frac{16}{5}$

$D = \sqrt[]{{(\frac{- 16}{5})}^{2} + {(\frac{8}{5})}^{2}}$

if 2 - = 16

$D = \frac{16}{\sqrt[]{5}}$

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New answer posted

2 months ago

Straight Lines Class 11th

Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is:

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Vishal Baghel

Contributor-Level 10

Equation of $⊥^{r}$ bisector of

$A B : y - 3 = \frac{t}{3} (x - t)$

For C put x = 0 so C $(0, 3 - \frac{t^{2}}{3})$

$h = \frac{t}{2} & K = \frac{6 - \frac{t^{2}}{3}}{2}$

$2 k = 6 - \frac{1}{3} * 4 h^{2}$

$2 x^{2} + 3 y - 9 = 0$

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2 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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2 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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2 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

2 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

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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

2 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

2 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

2 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

2 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:

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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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