Let m1, m2 be the slopes of two adjacent sides of a square of side a such that a2 + 11a + <math><mrow><mn>3</mn><mrow><mo>(</mo><mrow><msubsup><mrow><mi>m</mi></mrow><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></msubsup><mo>+</mo><msubsup><mrow><mi>m</mi></mrow><mrow><mn>2</mn></mrow><mrow><mn>2</mn></mrow></msubsup></mrow><mo>)</mo></mrow><mo>=</mo><mn>2</mn><mn>2</mn><mn>0</mn><mo>.</mo></mrow></math> If one vertex of the square is <math><mrow><mrow><mo>(</mo><mrow><mn>1</mn><mn>0</mn><mrow><mo>(</mo><mrow><mi>c</mi><mi>o</mi><mi>s</mi><mi>α</mi><mo>−</mo><mi>s</mi><mi>i</mi><mi>n</mi><mi>α</mi></mrow><mo>)</mo></mrow><mo>,</mo><mn>1</mn><mn>0</mn><mrow><mo>(</mo><mrow><mi>s</mi><mi>i</mi><mi>n</mi><mi>α</mi><mo>+</mo><mi>c</mi><mi>o</mi><mi>s</mi><mi>α</mi></mrow><mo>)</mo></mrow></mrow><mo>)</mo></mrow><mo>,</mo></mrow></math> where <math><mrow><mi>α</mi><mo>∈</mo><mrow><mo>(</mo><mrow><mn>0</mn><mo>,</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow></math> and the equation of one diagonal is <math><mrow><mrow><mo>(</mo><mrow><mi>c</mi><mi>o</mi><mi>s</mi><mi>α</mi><mo>−</mo><mi>s</mi><mi>i</mi><mi>n</mi><mi>α</mi></mrow><mo>)</mo></mrow><mi>x</mi><mo>+</mo><mrow><mo>(</mo><mrow><mi>s</mi><mi>i</mi><mi>n</mi><mi>α</mi><mo>+</mo><mi>c</mi><mi>o</mi><mi>s</mi><mi>α</mi></mrow><mo>)</mo></mrow><mi>y</mi><mo>=</mo><mn>1</mn><mn>0</mn><mo>,</mo></mrow></math> then <math><mrow><mn>7</mn><mn>2</mn><mrow><mo>(</mo><mrow><mi>s</mi><mi>i</mi><msup><mrow><mi>n</mi></mrow><mrow><mn>4</mn></mrow></msup><mi>α</mi><mo>+</mo><mi>c</mi><mi>o</mi><msup><mrow><mi>s</mi></mrow><mrow><mn>4</mn></mrow></msup><mi>α</mi></mrow><mo>)</mo></mrow><mo>+</mo><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>3</mn><mi>a</mi><mo>+</mo><mn>1</mn><mn>3</mn></mrow></math> is equal to

m1m2=−1, for square a,b,c,d letA(10(cosα−sinα),10(sinα+cosα))Diagonal : (cosα - sinα)x + (sinα + cosα)y = 10BD (diagonal)Dist. Of BD from A is|10(cosα−sinα)2+10(sinα+cosα)2−10|2=a2102=a2⇒a=10Also, a2 + 11a + 3 (m12+m22)=220210 + 3 (cm12+m22)=220m12+m22=103Also, m1 m2 = -1m2 + 1m2=103or −3,13m = 3,−13m4−103m2+1=0⇒m2=103±1009−42−103±832=3,13m = ±3,±13Diagonal AC:(sinα+cosα)x−(cosα−sinα)y=10 cos2α - 10cos2α = 0Slope of AC = sinα+cosαcosα−sinα=tanα+11−tanα=tan(α+π4)α=30°FIGURE? = 72(116+916)+100−30+13=72016+83=128

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Straight Lines Ncert Solutions Maths class 11th Maths 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

Option 1 -

119

Option 2 -

128

Option 3 -

145

Option 4 -

155

6 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$m_{1} m_{2} = - 1,$ 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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