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The perpendicular bisector of the line segment joining P(1,4)\; and\; Q(k,3) has y- intercept – 4 . Then a possible value of k is

  • Option 1)

    – 4

  • Option 2)


  • Option 3)


  • Option 4)

    – 2


Answers (1)


As we learnt in

Slope of a line -

If \Theta is the angle at which a straight line is inclined to a positive direction of x-axis, then the slope is defined by

m= \tan \Theta.

- wherein




Mid-point formula -

x= \frac{x_{1}+x_{2}}{2}

y= \frac{y_{1}+y_{2}}{2}


- wherein

If the point P(x,y) is the mid point of line joining A(x1,y1) and B(x2,y2) .


 As well as,


Slope – point from of a straight line -


- wherein

m\rightarrow slope

\left ( x_{1},y_{1} \right )\rightarrow point through which line passes


Mid-point of PQ    is    {R}\left ( \frac{k+1}{2}\:,\:\frac{7}{2} \right )


Slope of    PQ    is \frac{1}{1-k}


Slope of line perpendicular to    PQ=(k-1)

\left ( y-\frac{7}{2} \right )=(k-1)\left( x- \left( \frac{k+1}{2} \right ) \right )

y- intercept=-4, so point is (0,-4)

 \left ( -4-\frac{7}{2} \right )=(k-1)\left( - \left( \frac{k+1}{2} \right ) \right )

\frac{15}{2}= \frac{(k-1)(k+1)}{2}

k^{2}=16\:\:\:\Rightarrow k=\pm 4


Option 1)

– 4

This option is correct.

Option 2)


This option is incorrect.

Option 3)


This option is incorrect.

Option 4)

– 2

This option is incorrect.

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