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.
Answers
Answered by
2
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:
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