The point of intersection of the lines y=‐3 and x=4 is ____
A) (‐3,4) B) (2,3) C) ‐2,2) D) (‐2,4)
Answers
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.