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The line joining (-1,1) and (2,-2) and the line joining (1,2) and (2, k) are perpendicular to each other for the following value of k:
- a)1
- b)0
- c)-1
- d)3
Correct answer is option 'D'. Can you explain this answer?
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The line joining (-1,1) and (2,-2) and the line joining (1,2) and (2, ...
Solution:
The slope of the line joining (-1,1) and (2,-2) is given by
$$m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-1}{2-(-1)}=-\frac{3}{3}=-1$$
Therefore, the equation of this line is
$$y-1=-1(x+1)$$
$$\Rightarrow x+y=0$$
Similarly, the slope of the line joining (1,2) and (2,k) is given by
$$m_2=\frac{y_2-y_1}{x_2-x_1}=\frac{k-2}{2-1}=k-2$$
Therefore, the equation of this line is
$$y-2=(k-2)(x-1)$$
$$\Rightarrow y=kx+(-k+4)$$
Now, as these two lines are perpendicular to each other, their slopes are negative reciprocals of each other. Therefore,
$$m_1 \cdot m_2 = -1$$
$$\Rightarrow -1 \cdot (k-2)=-1$$
$$\Rightarrow k-2=1$$
$$\Rightarrow k=3$$
Therefore, the value of k for which the two lines are perpendicular is 3. Thus, the correct option is (D).
The slope of the line joining (-1,1) and (2,-2) is given by
$$m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-1}{2-(-1)}=-\frac{3}{3}=-1$$
Therefore, the equation of this line is
$$y-1=-1(x+1)$$
$$\Rightarrow x+y=0$$
Similarly, the slope of the line joining (1,2) and (2,k) is given by
$$m_2=\frac{y_2-y_1}{x_2-x_1}=\frac{k-2}{2-1}=k-2$$
Therefore, the equation of this line is
$$y-2=(k-2)(x-1)$$
$$\Rightarrow y=kx+(-k+4)$$
Now, as these two lines are perpendicular to each other, their slopes are negative reciprocals of each other. Therefore,
$$m_1 \cdot m_2 = -1$$
$$\Rightarrow -1 \cdot (k-2)=-1$$
$$\Rightarrow k-2=1$$
$$\Rightarrow k=3$$
Therefore, the value of k for which the two lines are perpendicular is 3. Thus, the correct option is (D).
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The line joining (-1,1) and (2,-2) and the line joining (1,2) and (2, k) are perpendicular to each other for the following value of k:a)1b)0c)-1d)3Correct answer is option 'D'. Can you explain this answer?
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