the inclination of the straight line passing through the point ( -3,6) and the midpoint of the line joining the point (4,-5) and (-2,9) is

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Solution

The mid-point of a line joining the points (4,-5) and (-2,9) is =

$(\frac{4 + (- 2)}{2}, \frac{(- 5) + 9}{2}) = (1, 2) N o w i n c l i n a t i o n o f t h e l i n e p a s \sin g t h r o u g h (- 3, 6) a n d (1, 2) i s g i v e n b y : - \tan θ = \frac{y_{2} - y_{1}}{x 2 - x_{1}} = \frac{2 - 6}{1 - (- 3)} = \frac{- 4}{4} = - 1 θ = 135^{o}$

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The inclination of the straight line passing through the point $(- 3, 6)$ and the mid-point of the line joining the point $(4, - 5)$ and $(- 2, 9)$ is

Find out the inclination of the straight line passing through the point (-3,6) and mid point of the line joining (4,-5) and (-2,9).

Q. The inclination of the straight line passing through the point (−3, 6) and the mid-point of the line joining the point (4, −5) and (−2, 9) is
(a) π/4
(b) π/6
(c) π/3
(d) 3 π/4
(e) 5 π/6

Q. Find the equation of the line passing through the point (−3, 5) and perpendicular to the line joining (2, 5) and (−3, 6).

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the inclination of the straight line passing through the point ( -3,6) and the midpoint of the line joining the point (4,-5) and (-2,9) is

The mid-point of a line joining the points (4,-5) and (-2,9) is = 4+-22,-5+92=(1,2) Now inclination of the line passing through (-3,6) and (1,2) is given by:-tanθ=y2-y1x2-x1=2-61-(-3)=-44=-1θ=135o