Question
The line passing through the points 
, (3,0) is
Hint:
There are terms in both x and y in the equation of a straight line. If the equation of the line is satisfied at point P(x,y), then point P is on line L. Y = a, where an is the y-coordinate of the line's points, is the equation for lines that are horizontal or parallel to the X-axis. Here we have to find the line passing through the points , (3,0).
The correct answer is:
x = a, where an is the x-coordinate of each point on the line, is the equation of a straight line that is vertical or parallel to the Y-axis.
For instance, the equation of the line passing through the point (2,3) and parallel to the X-axis is y= 3.
The line that is perpendicular to the Y-axis and contains the point (3,4) also has the equation x = 3.
Here we have given the points , (3,0).
Now, the equation of line joining the points (r1,θ1) and (r2,θ2) is given by:
So here we used the concept of the equation of the line passing through two points. Here we also used the trigonometric terms to find the answer using the formulas. So the final solution is 
.
Related Questions to study
Statement-I : If 
then A=
Statement-II : If 
then 
Which of the above statements is true
Statement-I : If then A=
Statement-II : If then
Which of the above statements is true
If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, one of the roots will be in the interval of (−α,a) and the other root will be in the interval (b,α).
If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, one of the roots will be in the interval of (−α,a) and the other root will be in the interval (b,α).
If 
be that roots 
where 
, such that 
and 
then the number of integral solutions of λ is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the number of integral solutions of λ is in between 
If be that roots where , such that and then the number of integral solutions of λ is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the number of integral solutions of λ is in between
If α,β then the equation
with roots 
will be
Here we used the concept of quadratic equations and solved the problem. We found the sum and product of the roots first and then proceeded for the final answer. Therefore, 
will be the equation for the roots .
If α,β then the equation with roots will be
Here we used the concept of quadratic equations and solved the problem. We found the sum and product of the roots first and then proceeded for the final answer. Therefore, will be the equation for the roots .
The equation of the directrix of the conic 
is
The equation of the directrix of the conic is
The conic with length of latus rectum 6 and eccentricity is
The conic with length of latus rectum 6 and eccentricity is
For the circle 
centre and radius are
For the circle centre and radius are
The polar equation of the circle of radius 5 and touching the initial line at the pole is
The polar equation of the circle of radius 5 and touching the initial line at the pole is
The circle with centre at and radius 2 is
The circle with centre at and radius 2 is
Statement-I : If 
Statement-II :If 
Which of the above statements is true
Statement-I : If
Statement-II :If
Which of the above statements is true
Then
R, let [x] denote the greatest integer 
x, then value of
+
+ 
+…+
is -
