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

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

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