A line is such that it’s segment between the lines. 5x-y+4=0and3x+4y-4=o is bisected at the point (1,5)​

Question

A line is such that it’s segment between the lines. 5x-y+4=0and3x+4y-4=o is bisected at the point (1,5)​

in progress 0
Aubrey 2 weeks 2021-09-10T10:34:33+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-09-10T10:35:46+00:00

    Answer:

    107x-3y-92=0

    Step-by-step explanation:

    What would you like to ask?

    MATHS

    A line is such that its segment between the straight lines 5x−y+4=0 and 3x+4y−4=0 is bisected at the point (1,5). Obtain its equation.

    Share

    Study later

    ANSWER

    Given lines are,

    5x−y+4=0 …… (1)

    3x+4y−4=0 …… (2)

    Let AB be the segment between the lines (1) and (2) and point P (1,5) be the mid-point of AB.

    Let the points be A(α

    1

    1

    ).

    It is given that the line segment AB is bisected at the point P (1,5). Therefore,

    ⇒(1,5)=(

    2

    α

    1

    2

    ,

    2

    β

    1

    2

    )

    So,

    2

    α

    1

    2

    =1

    α

    1

    2

    =2

    α

    2

    =2−α

    1

    …… (3)

    Also,

    2

    β

    1

    2

    =5

    β

    1

    2

    =10

    β

    2

    =10−β

    1

    …… (4)

    Point A and B lie on lines (1) and (2), respectively. Therefore, from lines (1) and (2), we have

    1

    −β

    1

    +4=0 …… (5)

    2

    +4β

    2

    −4=0 ……. (6)

    Now, from equations (3) and (4) and equations (5) and (6), we have,

    3(2−α

    1

    )+4(10−β

    1

    )−4=0

    6−3α

    1

    +40−4β

    1

    −4=0

    −3α

    1

    −4β

    1

    +42=0

    1

    +4β

    1

    −42=0 …… (7)

    Now, from equations (5) and (7), we have

    α

    1

    =

    23

    26

    , β

    1

    =

    23

    222

    Thus, the line is passing through A(

    23

    26

    ,

    23

    222

    ) and P (1,5). Therefore, the equation of the line is,

    y−5=

    23

    26

    −1

    23

    222

    −5

    (x−1)

    y−5=

    23

    26−23

    23

    222−115

    (x−1)

    y−5=

    3

    107

    (x−1)

    3y−15=107x−107

    107x−3y−92=0

    Hence, this is the required equation.

    solution

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )