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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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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

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  • Distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $ \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Let $y=mx+c$ be the line through the point A(-1, 2)
$ \therefore 2 = m(-1)+c$
$ \Rightarrow 2 = -m+c$
or $c=m+2$
Hence $y=mx+m+2$------(1)
Equation of the given line is
Let us solve equation (1) and (2) for $x$ and $y$.
$ x + mx+m+2=4$
$ \Rightarrow x(1+m)=2-m$
$ \therefore x = \large\frac{2-m}{1+m}$
also $ y = \large\frac{5m+2}{1+m}$
Hence $B \bigg( \large\frac{2-m}{1+m}$$, \large\frac{5m+2}{1+m} \bigg)$ is the point of intersection.
It is given that the distance between the two points A and B is 3 units.
$ \therefore \overline{AB} = \sqrt{ \bigg( \large\frac{2-m}{1+m}+1 \bigg)^2 + \bigg( \large\frac{5m+2}{1+m}-2 \bigg)^2}$
$ \Rightarrow 3 = \sqrt{ \bigg[ \large\frac{2-m+1+m}{1+m} \bigg]^2+ \bigg[ \large\frac{5m+2-2-2m}{1+m} \bigg]^2}$
Squaring on both sides we get,
$9= \large\frac{9}{(1+m)^2}$$ + \large\frac{9m^2}{(1+m)^2}$
$ \Rightarrow 1 = \large\frac{1}{(1+m)^2}$$+ \large\frac{m^2}{(1+m)^2}$
$ \Rightarrow (1+m)^2= 1+m^2$
i.e., $1+2m+m^2=1+m^2$
$ \Rightarrow m = 0$
Hence the slope of the required line is 0.
$ \therefore $ The line must be parallel to $x$ - axis.
answered May 21, 2014 by thanvigandhi_1

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