Question

Solve Q no 10 and 11 c
Ans=10 3x+4y+=13
11 c 2y=x-5,(0,-5/2)
Its neccessary for my exam

Answer 1

10. ans - eqn of line = 4x - 3y + 12 = 0
putting y = 0 in above eqn we get,
x = -3 so coordinates of A = (-3,0)
now slope of this line = 4/3
now let the slope of required line be m
then as required line is perpendicular to given line so m.(4/3) = -1
solving above eqn we get m = -3/4
so eqn of required line => y - 0 = -3/4(x -(-3))
=> 4y = -3x - 9
=> 3x + 4y + 9 = 0

Answer 2

Equation of straight line:  L : 4x - 3 y + 12 = 0
Point A = Intersection of X-axis and L:     So  y = 0.
     
    So    4 x - 3*0 + 12 = 0     =>   x = -12/4 = -3.

 => Point A = (-3, 0).

Line L :  4 x - 3 y + 12 = 0
         or,  3 y = 4x + 12    or     y = 4 x /3  + 4
Slope of the Line L : m = 4/3.

Slope of any line L' perpendicular to line L = - 1/m = -3/4.

Equation of a Line L' with slope -3/4 and passing thru A(-3,0) is:

              (y - 0) / (x - (-3)) = -3/4
               y /(x+3) = - 3/4
               4 y = -3x - 9
               4 y + 3x + 9 = 0...     ANSWER.

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11. C.

Given Line L :  4 y = 2 x + 3
=>   Slope of L :  m = 2/4 =1/2

Equation of a line L' (parallel to L) with x intercept = 5, and  y intercept = b:

                  x / 5 + y / b = 1      or,     5 y + b x = 5 b

So its slope = - b / 5 = m = 1/2
                => b = - 5/2

So the equation of line L' :   5 y + (-5/2) x = 5(-5/2)
                        or,                    2 y - x + 5 = 0.

     L' cuts the y-axis at : (0, b) = (0, -5/2)