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# Can someone explain - Co-ordinate geometry - JEE Main-4

The perpendicular bisector of the line segment joining $\dpi{100} P(1,4)\; and\; Q(k,3)$ has $\dpi{100} y-$ intercept – 4 . Then a possible value of $\dpi{100} k$ is

• Option 1)

– 4

• Option 2)

1

• Option 3)

2

• Option 4)

– 2

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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

and,

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 -

$y-y_{1}=m(x-x_{1})$

- 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)

1

This option is incorrect.

Option 3)

2

This option is incorrect.

Option 4)

– 2

This option is incorrect.

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