Question

Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.

Answer 1

The equation of line is x+y=4. The distance of the line from the point of intersection is 3 units.

Let y=mx+c be the equation of line passing through the point ( −1,2 ).

The point ( −1,2 ) satisfies the equation of the line.

2=m×−1+c c=m+2

Substitute the value of c in equation y=mx+c.

y=mx+m+2(1)

The other equation of line is,

x+y=4(2)

Substitute the value of y=4−x in equation (1).

4−x=mx+m+2 4−m−2=mx+x 2−m=( m+1 )x x= 2−m ( m+1 )

Substitute the value of x in equation (2) to obtain the value of y.

y=4− 2−m ( m+1 ) = 4m+4−2+m m+1 = 5m+2 m+1

The coordinates of the point of intersection is ( 2−m m+1 , 5m+2 m+1 ).

This point is at a distance of 3 units from point ( −1,2 ).

The formula for the distance between two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by,

d= ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2

Substitute the values of d=3, ( x 1 , y 1 ) and ( x 2 , y 2 ) as ( −1,2 ) and ( 2−m m+1 , 5m+2 m+1 ) in the above expression.

3= ( 2−m m+1 +1 ) 2 + ( 5m+2 m+1 −2 ) 2 = ( 2−m+1×( m+1 ) m+1 ) 2 + ( 5m+2−2×( m+1 ) m+1 ) 2 = ( 2−m+m+1 m+1 ) 2 + ( 5m+2−2m−2 m+1 ) 2 = ( 3 m+1 ) 2 + ( 3m m+1 ) 2

Further simplify the above expression.

9= ( 3 m+1 ) 2 + ( 3m m+1 ) 2 9= 9 ( m+1 ) 2 + 9 m 2 ( m+1 ) 2 9= 9( m 2 +1 ) m 2 +1+2m m 2 +1+2m= m 2 +1

Further simplify the above expression.

2m=0 m=0

Thus, the slope of line is 0. The line must be drawn parallel to the x axis.