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)
Answers
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
,β
1
).
It is given that the line segment AB is bisected at the point P (1,5). Therefore,
⇒(1,5)=(
2
α
1
+α
2
,
2
β
1
+β
2
)
So,
2
α
1
+α
2
=1
α
1
+α
2
=2
α
2
=2−α
1
...... (3)
Also,
2
β
1
+β
2
=5
β
1
+β
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
5α
1
−β
1
+4=0 ...... (5)
3α
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
3α
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