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.
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
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