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Question

The line passing through the points open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses, (3,0) is

  1. r left square bracket 2 S i n space theta plus 3 C o s space theta right square bracket equals 6    
  2. r left square bracket 2 S i n space theta plus 3 C o s space theta right square bracket equals 5    
  3. r left square bracket 2 C o s space theta plus 3 s i n space theta right square bracket equals 6    
  4. r left square bracket negative 3 s i n space theta plus 2 C o s space theta right square bracket equals 5    

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 open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses, (3,0).

The correct answer is: r left square bracket 2 S i n space theta plus 3 C o s space theta right square bracket equals 5


    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  open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses, (3,0).
    Now, the equation of line joining the points (r1,θ1) and (r2,θ2) is given by:
    fraction numerator sin left parenthesis theta 2 minus theta 1 right parenthesis over denominator r end fraction equals fraction numerator sin left parenthesis theta minus theta 2 right parenthesis over denominator r 1 end fraction equals fraction numerator sin left parenthesis theta 1 minus theta right parenthesis over denominator r 2 end fraction
N o w space p u t t i n g space t h e space g i v e n space p o i n t s space i n space t h i s comma space w e space g e t colon
fraction numerator sin left parenthesis 0 minus begin display style straight pi over 2 end style right parenthesis over denominator r end fraction equals fraction numerator sin left parenthesis theta minus 0 right parenthesis over denominator 2 end fraction equals fraction numerator sin left parenthesis begin display style straight pi over 2 end style minus theta right parenthesis over denominator 3 end fraction
fraction numerator negative 1 over denominator r end fraction equals fraction numerator sin left parenthesis theta right parenthesis over denominator 2 end fraction equals fraction numerator cos left parenthesis theta right parenthesis over denominator 3 end fraction
N o w space e q u a t i n g space t h e space t e r m s comma space w e space g e t colon
3 sin theta plus 2 cos theta equals fraction numerator negative 1 cross times 2 cross times 3 over denominator r end fraction
3 sin theta plus 2 cos theta equals fraction numerator negative 6 over denominator r end fraction
r left square bracket 3 sin theta plus 2 cos theta right square bracket equals negative 6

    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 r left square bracket 3 s i n space theta plus 2 C o s space theta right square bracket equals negative 6.

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