If the perpendicular bisector of the line seg | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the points be $A(\alpha, 3)$ and $B(2, -1)$. The midpoint $M$ of $AB$ is $\left(\frac{\alpha+2}{2}, \frac{3-1}{2}\right) = \left(\frac{\alpha+

Vedclass · Accepted Answer

$\pm 2$

Vedclass · Answer

(C) Let the points be $A(\alpha, 3)$ and $B(2, -1)$. The midpoint $M$ of $AB$ is $\left(\frac{\alpha+2}{2}, \frac{3-1}{2}ight) = \left(\frac{\alpha+2}{2}, 1ight)$.Since the perpendicular bisector passes through $M$ and has a $y$-intercept of $1$,the point $M$ must lie on the $y$-axis or the line must be horizontal. However,the $y$-intercept is $1$,which is the same as the $y$-coordinate of $M$. This implies the $x$-coordinate of $M$ must be $0$.$\frac{\alpha+2}{2} = 0 \implies \alpha = -2$.Alternatively,the slope of $AB$ is $m_{AB} = \frac{-1-3}{2-\alpha} = \frac{-4}{2-\alpha} = \frac{4}{\alpha-2}$.The slope of the perpendicular bisector is $m = -\frac{1}{m_{AB}} = \frac{2-\alpha}{4}$.The equation of the line is $y - 1 = m(x - \frac{\alpha+2}{2})$.Given the $y$-intercept is $1$,we set $x=0$ and $y=1$:$1 - 1 = m(0 - \frac{\alpha+2}{2}) \implies 0 = m(-\frac{\alpha+2}{2})$.Since $m 
eq 0$ (unless $\alpha=2$,which makes $AB$ vertical),we must have $\alpha+2 = 0$,so $\alpha = -2$.Wait,checking the case where the line is horizontal: if $m=0$,then $\alpha=2$. If $\alpha=2$,$A=(2,3)$ and $B=(2,-1)$. The perpendicular bisector is $y=1$,which has $y$-intercept $1$. Thus $\alpha=2$ is also a solution.Therefore,$\alpha = \pm 2$.

If the perpendicular bisector of the line segment joining $A(\alpha, 3)$ and $B(2, -1)$ has $y$-intercept $1$,then $\alpha =$

Similar Questions

$A$ line passing through $P(2,3)$ and making an angle of $30^{\circ}$ with the positive direction of the $x$-axis meets the curve $x^2-2xy-y^2=0$ at $A$ and $B$. Then the value of $PA \cdot PB$ is

If $P(x_1, y_1)$ and $Q(x_2, y_2)$ are points on $2x + 3y + 1 = 0$ such that $|PA - PB|$ is maximum and $|QA - QB|$ is minimum,where $A(2, 0)$ and $B(0, 2)$,then the value of $x_1 - y_1 + x_2 - y_2$ is -

The number of points having a distance of $\sqrt{5}$ from the straight line $x-2y+1=0$ and a distance of $\sqrt{13}$ from the line $2x+3y-1=0$ is:

The equation of one of the sides of an isosceles right-angled triangle,whose hypotenuse is $3x + 4y = 4$ and the vertex opposite to the hypotenuse is $(2, 2)$,is:

The equation of the straight line perpendicular to $5x - 2y = 7$ and passing through the point of intersection of the lines $2x + 3y = 1$ and $3x + 4y = 6$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module