Question

Find the slope of line AB with the points lying on
1. A(2, 1), B(2, 6)
2. A(−4, 2), B(−4, −2)
3. A(−2, 8), B(−2, −2)
Justify that the line segment formed by given points is parallel to Y-axis. What can you say about their slope? Why?

Answer 1

concepts :-
1. slope of line joining points $(x_1,y_1)$ and $(x_2,y_2)$ is given as
$m=\frac{y_2-y_1}{x_2-x_1}$ and equation of line $(y-y_1)=m(x-x_1)$

2. any line is parallel to Y - axis only when slope of line = ∞

(i) A(2,1) and B(2,6)
slope of AB = (6 - 1)/(2 - 2) = 5/0 = ∞
hence, line AB is parallel to Y -axis .

(ii) A(-4,2) and B(-4,-2)
slope of line AB = (-2 - 2)/(-4 + 4) = -4/0 = ∞
hence, line AB is parallel to Y-axis.

(iii) A(-2,8) and B(-2,-2)
slope of line AB = (-2 -8)/(-2 + 2) = -10/0 = ∞
hence, line AB is parallel to Y-axis .


all these equations are parallel to Y - axis. slope of these lines are undefined. means change in y with respect to x is undefined. we can't get real value of it.

Answer 2

Answer:

=> {cosθ(1 + sinθ) + cosθ(1 - sinθ)}/(1 - sinθ)(1 + sinθ) = 4

=> {cosθ + cosθ.sinθ + cosθ - cosθ.sinθ}/(1 - sin²θ) = 4

=> 2cosθ/cos²θ = 4 [ we know, sin²x + cos²x = 1 so, (1 - sin²θ) = cos²θ]

=> 2/cosθ = 4

=> cosθ = 1/2 = cos60°

hence, in 0 < θ < 90° , θ = 60°

now, if given equation is not defined.

(1 - sinθ) = 0

in 0 < θ < 90° , sinθ = 1 at 90°

hence, equation is undefined at θ = 90°

[ note : one more case for undefined, (1 + sinθ) = 0 , but in 0 < θ < 90° it's not possible. thars why I didn't mention it above]

Step-by-step explanation: