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Find the distance between the given line and point.
1. L1: 4x + 3y = 5 and P(2, 0)
2. L2: x =7y - 3 and P(1, -1)
3. L3: 3y = 2x + 1 and P(2, -4)
4. L4: 2x = 10 - 5y and P(5, 0)
5. L5: 3y - x = x - 1 and P(3, -2)
6. L6: 5x = 1 and P(-1, -4)
7. L7: 6y = 3y +2 and P(2, 1)
8. L8: 3x = 10 – y and P(1, 7)
9. L9: y = 2x + 7 and P(2, 3)
10. L10: y - 4x = 3x - y +1 and P(-2, -3)​


Sagot :

Answer:

Distance = |ax1 + by1 + c| / √(a^2 + b^2)

1. L1: 4x + 3y = 5 and P(2, 0)

Distance = |4(2) + 3(0) - 5| / √(4^2 + 3^2)

= |8 - 5| / √(16 + 9)

= 3 / √25

= 3 / 5

2. L2: x = 7y - 3 and P(1, -1)

Distance = |1 - 7(-1) + 3| / √(1^2 + 7^2)

= |1 + 7 + 3| / √(1 + 49)

= 11 / √50

= 11 / (5√2)

3. L3: 3y = 2x + 1 and P(2, -4)

Distance = |2(2) - 3(-4) - 1| / √(2^2 + 3^2)

= |4 + 12 - 1| / √(4 + 9)

= 15 / √13

4. L4: 2x = 10 - 5y and P(5, 0)

Distance = |2(5) - 10 - 0| / √(2^2 + 5^2)

= |10 - 10| / √(4 + 25)

= 0 / √29

= 0

5. L5: 3y - x = x - 1 and P(3, -2)

Distance = |3(-2) - 3 - 1| / √(3^2 + 1^2)

= |-6 - 4| / √(9 + 1)

= 10 / √10

= 10 / (2√5)

6. L6: 5x = 1 and P(-1, -4)

Distance = |5(-1) - 1 + 4| / √(5^2)

= |-5 - 1 + 4| / √25

= 2 / 5

7. L7: 6y = 3y + 2 and P(2, 1)

Distance = |3(1) - 6 + 2| / √(3^2 + 6^2)

= |3 - 6 + 2| / √(9 + 36)

= |-1| / √45

= 1 / (3√5)

8. L8: 3x = 10 - y and P(1, 7)

Distance = |3(1) - 10 - 7| / √(3^2 + 1^2)

= |3 - 10 - 7| / √(9 + 1)

= |-14| / √10

= 14 / √10

9. L9: y = 2x + 7 and P(2, 3)

Distance = |2(2) - 3 + 7| / √(2^2 + 1^2)

= |4 - 3 + 7| / √(4 + 1)

= 8 / √5

10. L10: y - 4x = 3x - y + 1 and P(-2, -3)

Distance = |-3 - 4(-2) - 3 + 1| / √(4^2 + 1^2)

= |-3 + 8 - 3 + 1| / √(16 + 1)

= 3 / √17