find the equation of line passing through the point of intersection of the line x+y+1=0,2x-y+5=0 and the points are (5,-2)
Answers
Answer:
Given the lines x−y−6=0 and 2x+y+3=0
The point of intersection of the above two line is found by solving for x and
x−y−6=0−−−−−−−(1)
2x+y+3=0−−−−(2)
y=−5
x=1
P(1,−5)
Now the equation of the line is y−(−5)=1−(−5)2−1(x−1)
y+5=6x−6
6x−y−11=0
The two lines that intersect each other are given as:
x-y=6 …(1)
2x+y=-3 …(2)
Add (1) and (2), to get 3x = 3 or x = 1.
Put x=1 in (1) to get y = -5
So the two lines intersect at P (1,-5).
The equation of the line joining P and Q (2,1)
(x-1)/(2–1) = (y+5)/(1+5), or
(x-1)/1 = (y+5)/6, or
6x-6 = y+5, or
6x-y-11=0 is the equation of the line PQ.
Answer:
ANSWER
Point of intersections of the lines x−y+1=0 and 2x−3y+5=0 is obtained by solving the two equations simultaneously.
x−y+1=0 .....(1)
2x−3y+5=0 .....(2)
Equation (1)×3..........3x−3y+3=0
Equation (2)×1.........2x−3y+5=0
−+−
x−2=0
x=2
∴ y=3
equation of a line passing through (2,3) and having a slope 'm' is given by
⇒y−y
1
=m(x−x
1
)
y−3=m(x−2)
mx−y−2m+3=0
The above line is at a distance of
5
7
units from the point (3,2)
we know that the distance of a line ax+by+c=0 from (h,k) is given by
⇒d=
a
2
+b
2
ah+bk+c
Using the above formula we can write d=
5
7
⇒
1+m
2
3m−2−2m+3
=
5
7
⇒
1+m
2
m+1
=
5
7
squaring both sides we have ,
⇒25m
2
+25+50m=49+49m
2
⇒24m
2
−50m+24=0
⇒12m
2
−25m+12=0
⇒12m
2
−16m−9m+12=0
⇒4m(3m−4)−3(3m−4)=0
⇒(3m−4)(4m−3)=0
∴m=
3
4
or
4
3
∴ the equation of the line possible are
⇒
3
4
x−y−
3
8
−13=0 or
4
3
x−y−
4
6
+3=0
⇒4x−3y+1=0 or 3x−4y+6=0
Hence, the answer is 4x−3y+1=0 or 3x−4y+6=0.