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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.

Q: 16     Find the direction in which a straight line must be drawn through the point  \small (-1,2)  so that its point of intersection with the line  \small x+y=4  may be at a distance of 3 units from this point.

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Let (x_1,y_1) be the point of intersection
it lies on line \small x+y=4 
Therefore, 
x_1+y_1=4 \\ x_1=4-y_1\ \ \ \ \ \ \ \ \ \ \ -(i)
Distance of point (x_1,y_1) from \small (-1,2) is 3
Therefore,
3= |\sqrt{(x_1+1)^2+(y_1-2)^2}|
Square both the sides and put value from equation (i)
9= (5-y_1)^2+(y_1-2)^2\\ 9=y_1^2+25-10y_1+y_1^2+4-4y_1\\ 2y_1^2-14y_1+20=0\\ y_1^2-7y_1+10=0\\ y_1^2-5y_1-2y_1+10=0\\ (y_1-2)(y_1-5)=0\\ y_1=2 \ or \ y_1 = 5
When y_1 = 2 \Rightarrow x_1 = 2       point is  (2,2)
and
When y_1 = 5 \Rightarrow x_1 = -1      point is (-1,5)
Now, slope of line joining point  (2,2)  and  \small (-1,2)  is 
m = \frac{2-2}{-1-2}=0
Therefore, line is parallel to x-axis                       -(i)

or
slope of line joining point (-1,5)  and \small (-1,2) 
m = \frac{5-2}{-1+2}=\infty
Therefore, line is parallel to y-axis                      -(ii)

Therefore, line is parallel to x -axis or parallel to y-axis

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