# 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.

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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