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.

Asked by Abhisek | 1 year ago |  103

##### Solution :-

Consider y = mx + c as the line passing through the point (-1, 2)

So we get

2 = m (-1) + c

By further calculation

2 = -m + c

c = m + 2

Substituting the value of c

y = mx + m + 2 …… (1)

So the given line is

x + y = 4 ……. (2)

By solving both the equations we get

$$x=\dfrac{2-m}{m+1}$$ and $$y=\dfrac{5m+2}{m+1}$$

$$\dfrac{2-m}{m+1},\dfrac{5m+2}{m+1}$$ is the point of intersection of lines (1) and (2) .

Given that, the point is at a distance of 3 units from (-1,2)

By distance formula,

$$\sqrt{ (\dfrac{2-m}{m+1}+1)^2+(\dfrac{5m+2}{m+1}-2)^2}=3$$

Square both sides,

$$(\dfrac{2-m+m+1}{m+1})^2+(\dfrac{5m+2-2m-2}{m+1})^2=3^2$$

$$\dfrac{9}{(m+1)^2}+\dfrac{9m^2}{(m+1)^2}=9$$

Divide the equation by 9 ,

$$\dfrac{1+m^2}{(m+1)^2}=1$$

By cross multiplication

1 + m2 = m2 + 1 + 2m

So we get

2m = 0

m = 0

Therefore, the slope of the required line must be zero that is, the line must be parallel to the x-axis.

Answered by Pragya Singh | 1 year ago

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