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 is parallel to x-axis and y-axis.
how to do it

Answer 1

Answer:

Step-by-step explanation:

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

∴2=m(−1)+c

⇒2=−m+c

or c=m+2

Hence y=mx+m+2------(1)

Equation of the given line is

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

Let us solve equation (1) and (2) for x and y.

x+mx+m+2=4  

⇒x (1+m)=2−m

∴x= 2−m/1+m

also y=5m+2/1+m

Hence B(2−m/1+m , 5m+2/1+m) is the point of intersection.

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

∴AB = √([2−m/1+m] + 1)² + ([5m+2/1+m] − 2)²

⇒3 = √([2−m/1+m] + 1)² + ([5m+2/1+m] − 2)²

Squaring on both sides we get,

9 = 9/(1+m)² + 9m²/(1+m)²

⇒1 = 1 / (1+m)²+ m²/(1+m)²

⇒(1+m)²= 1+m²

i.e., 1 + 2m + m²=1 + m2

⇒m=0

Hence the slope of the required line is 0.

Therefore,  The line must be parallel to x - axis.