A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

We have to find the point of intersection of the lines

2x – 3y + 4 = 0

3x + 4y – 5 = 0  

[ 2x - 3y = (-4) ] x 3

[ 3x + 4y = 5 ] x 2

6x -9y = (-12)

6x +8y =10

subtracting

-17y =-22

 

y=(22/17)

substitute in 

3x + 4y = 5

3x = 5 -{88/17}=[-3/17]

x =(-1/17)

 

The person is standing at [ (-1/17) , (22/17) ] 

The shortest path is perpendicular to 6x – 7y + 8 = 0

Perpendicular line is 7x+6y+k=0

This passes through [ (-1/17) , (22/17) ] if

7 (-1/17) +6(22/17) +k =0

k = (-125/17)

Perpendicular line is 7x+6y+k=0

changes to  7x+6y+(-125/17)=0 

or 119x+102y-125=0 

119x+102y=125 is the required path.

 

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

 

ncert cbse chapter 9 sequences and series miscellaneous exercise

 

 32.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

solution

 

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