Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Match the questions given under Column C1 with their appropriate answers given under the Column C2 A) the coordinates of the points P and Q on the line x+5y=13 which are at a distance of 2 units from the line 12x-5y+26=0 are i) 3,1 minus 7 11

Match the questions given under Column C1 with their appropriate answers given under the Column C2.

Column C1  Column C2

a) The coordinates of the points P and Q on

the line  x+5y=13 which are at a distance of 2

units from the line 12x-5y+26=0 are

             i) (3,1), (-7,11)                           

b) The coordinates of the points on

the line  x+y=4 which are at a unit distance  from the line 4x-3y+10=0 are

   ii) -1/3, 11/3, 4/3, 7/3

c) The coordinates of the points on the line  joining A 

(-2,5) and B(3,1) such that AP=+Q=QB are

 iii) 1, 12/5,-3,16/5

 

 

 

Answers (1)

Let P (x1,y1) be any point on the given line x+5y=13 

  x1+5y1=13   

 5y1=13-x1….(i)  

Distance of the point P(x1,y1)from the equation 12x-5y+26=0  

d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B ^{2}}}

2=\frac{\left | 12x_{1}+5y_{1}+26 \right |}{\sqrt{\left ( 12 \right )^{2}+\left ( -5 \right )^{2}}}

 2=\frac{\left | 12x_{1}-\left ( 13-x_{1} \right )+26 \right |}{\sqrt{144+25}} = \frac{\left | 12x_{1}-13+x_{1} +26 \right |}{13} 

  2=\frac{\left | 13x_{1}+13 \right |}{13}

 2=|x1+1| 

   2= ±(x1+1) 

 So x1=1….(ii) or x1=-3…. (iii)

    Putting the value in equation (i) we get    5y1=13-1=12

y_{1}=\frac{12}{5} 

   Putting the value of x1=-3  in the same equation we get 5y1=13-(-3)=16 

y_{1}=\frac{16}{5}

  Hence, the required points on the given line are\left ( 1,\frac{12}{5} \right ) \, \, and\, \, \left ( -3,\frac{16}{5} \right )and -3,165 

 Hence, (a)-(iii)

    bLet P x1, y1 be any point lying in the equation x+y=4   

 x1+y1=4….  i

  Now, the distance of the point from the equation is   d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+ B ^{2}}}

1=\frac{\left | 4x_{1}+3y_{1}-10 \right |}{\sqrt{(4)^{2}+(3)2}}=\frac{\left | 4x_{1}+3y_{1}-10 \right |}{\sqrt{16-9}}  

 1=\left |\frac{ 4x_{1}+3y_{1}-10 }{5}\right |

   4x1+3y1-10= ±5 

  either  4x1+3y1-10=5   or 4x1+3y1-10=-5  

   4x1+3y1=15 ….(ii) or   4x1+3y1=5…..(iii)

  From equation i we have y1=4-x1….(iv) 

 Putting the value of y1 in equation ii we get  4x1+34-x1=15  

  4x1+12-3x1=15   

 x1=3   Putting the value of x1 in equation (iv) we get  y1=4-3=1

  Putting the value of y1 in equation iii we get  4x1+34-x1=5 

  4x1+12-3x1=5 

   x1=5-12=-7  

Putting the value of x1 in equation (iv) , we get  y1=4-(-7) 

  y1=4+7=11 

  Hence, the required points on the given line are (3,1) and (-7,11) 

 Hence, b-i   

c  Given that AP=PQ=QB  and given points are A(-2,5) and B(3,1) 

 Firstly, we find the slope of the line joining the points (-2,5) and (3,1)  

 Slope of line joining two points=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

m_{AB}=\frac{1-5}{3-\left ( -2 \right )}=-\frac{4}{3+2}=-\frac{4}{5}

  Now equation of line passing through the point (-2, 5) y-5=-4/5[x-(-2)]  

 5y-25=-4(x+2) 

 4x+5y-17=0  Let P (x1,y1) and Q (x2,y2) be any two points on the AB 

  P(x1,y1) divides the line AB in the ratio 1:2  

x_{1}=\frac{1*3+2*(-2)}{1+2}=\frac{3-4}{3}=\frac{1}{3}

y_{1}=\frac{1*1+2*5}{1+2}=\frac{1+10}{3}=\frac{11}{3}

  Now, Q (x2,y2) is the midpoint of PB  x_{2}=\frac{3+\left (-\frac{1}{3} \right )}{2}=\frac{8}{6}=\frac{4}{3}

y_{2}=\frac{1+{11}{3}}{2}=\frac{3+11}{6}=\frac{14}{6}=\frac{7}{3}

 Hence, the coordinates of Q (x2,y2) is (4/3,7/3) 

 Hence, c-ii

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question