Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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)
1

Option 3)
2

Option 4)
– 2

Answers (1)

best_answer

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.

Posted by

divya.saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year

anam-khan

JEE Mains: Farmer's son from remote Maharashtra village aces exam

Latest Question