Question

A straight line must be drawn through the point (-1, 2) so that its point

of Intersection with the line x + y = 4 may be at a distance of 3 units from

this point.

(i) Which straight line equation used for the given problem:

(a) Y- b = m (x – a) (b) y = mx + c (c) y -2x+3=0 (d) y= 3x-6

(ii) What is the point of intersection of two lines:

(a) (1, 2) (b) (-3, 4) (c) (

2−

+1

,

5+2

+1

)(d) (-7,-2)

(iii) Given distance 3 units for which two points :

(a) (1,2) and (

2−

+1

,

5+2

+1

) (b) (-7,-2) and (-3, 4) (c) (-1, 2) and

(1, 2) (d) (

2−

+1

,

5+2

+1

) and (-1, 2)

(iv) What is the slope(m) of the required line:

(a) m=0 (b) m =-1 (c) m = 2 (d) m = 4

(v) Which information we get from the given problem:

(a) A line must perpendicular to the x- axis (b) a line must parallel

to the x – axis (c) a line parallel to the y-axis (d) a line which

lies on the x- axis.

​

Answer 1

Answer:

ANSWER

sin2A=λsin2B

RHS

=

λ−1

λ+1

=

sin2B

sin2A

−1

sin2B

sin2A

+1

=

sin2A−sin2B

sin2A+sin2B

=

2sin(

2

2A−2B

)cos(

2

2A+2B

)

2sin(

2

2A−2B

)cos(

2

2A−2B

)

=

sin(A−B)cos(A+B)

sin(A+B)cos(A−B)

=

tan(A−B)

tan(A+B)

=LHS

Hence proved.

Step-by-step explanation:

jkhgo9gh! OK CNN jguvbkkbb...