Question

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)​

Answer 1

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.