Question

Find the direction in which a straight line should be drawn through a point a(-1,2) sobthat its point of intersection with the line x+y=4 may be at a distance of 3 units

Answer 1

Let y=mx+cy=mx+c be the line through the point A(-1, 2)

∴2=m(−1)+c∴2=m(−1)+c

⇒2=−m+c⇒2=−m+c

or c=m+2c=m+2

Hence y=mx+m+2y=mx+m+2------(1)

Equation of the given line is

x+y=4x+y=4------(2)

Let us solve equation (1) and (2) for xxand yy.

x+mx+m+2=4x+mx+m+2=4

⇒x(1+m)=2−m⇒x(1+m)=2−m

∴x=2−m1+m∴x=2−m1+m

also y=5m+21+my=5m+21+m

Hence B(2−m1+mB(2−m1+m,5m+21+m),5m+21+m) is the point of intersection.

It is given that the distance between the two points A and B is 3 units.

∴AB¯¯¯¯¯¯¯¯=(2−m1+m+1)2+(5m+21+m−2)2−−−−−−−−−−−−−−−−−−−−−−−−−−√∴AB¯=(2−m1+m+1)2+(5m+21+m−2)2

⇒3=[2−m+1+m1+m]2+[5m+2−2−2m1+m]2−−−−−−−−−−−−−−−⎷⇒3=[2−m+1+m1+m]2+[5m+2−2−2m1+m]2

Squaring on both sides we get,

9=9(1+m)29=9(1+m)2+9m2(1+m)2+9m2(1+m)2

⇒1=1(1+m)2⇒1=1(1+m)2+m2(1+m)2+m2(1+m)2

⇒(1+m)2=1+m2⇒(1+m)2=1+m2

i.e., 1+2m+m2=1+m21+2m+m2=1+m2

⇒m=0⇒m=0

Hence the slope of the required line is 0.

∴∴ The line must be parallel to xx - axis.