If the line 2x - 3y + 5 = 0 is the perpendicu | vedclass

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

Question

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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let the points be $A(1, -2)$ and $B(\alpha, \beta)$. The midpoint $M$ of $AB$ is $\left(\frac{\alpha + 1}{2}, \frac{\beta - 2}{2}ight)$.Since $M$ lies on the line $2x - 3y + 5 = 0$,we have $2\left(\frac{\alpha + 1}{2}ight) - 3\left(\frac{\beta - 2}{2}ight) + 5 = 0$,which simplifies to $2\alpha - 3\beta + 18 = 0$ $(i)$.The slope of line $AB$ is $m_1 = \frac{\beta + 2}{\alpha - 1}$. The slope of the given line $2x - 3y + 5 = 0$ is $m_2 = \frac{2}{3}$.Since $AB$ is perpendicular to the given line,$m_1 	imes m_2 = -1$,so $\left(\frac{\beta + 2}{\alpha - 1}ight) 	imes \frac{2}{3} = -1$,which gives $2\beta + 4 = -3\alpha + 3$,or $3\alpha + 2\beta + 1 = 0$ $(ii)$.Solving equations $(i)$ and $(ii)$ simultaneously: Multiply $(i)$ by $2$ and $(ii)$ by $3$: $4\alpha - 6\beta + 36 = 0$ and $9\alpha + 6\beta + 3 = 0$.Adding these gives $13\alpha + 39 = 0$,so $\alpha = -3$. Substituting $\alpha = -3$ into $(ii)$,we get $3(-3) + 2\beta + 1 = 0$,so $2\beta = 8$,which means $\beta = 4$.Thus,$\alpha + \beta = -3 + 4 = 1$.

If the line $2x - 3y + 5 = 0$ is the perpendicular bisector of the line segment joining $(1, -2)$ and $(\alpha, \beta)$,then $\alpha + \beta =$

Similar Questions

Find the equation of the line perpendicular to the line $ax + by + c = 0$ and passing through the point $(a, b)$.

If $\alpha$ and $\beta$ are the angles made by the normals drawn from the origin to the lines $x+y+\sqrt{2}=0$ and $x-\sqrt{3}y-2=0$ with the positive direction of the $X$-axis respectively,measured in the anti-clockwise direction,then $\alpha+\beta=$

The image of the point $(2, 4)$ with respect to the straight line $2x + 3y - 6 = 0$ is

What are the coordinates of the image of the point $(4, -5)$ with respect to the line $y = x$?

The perpendicular from the origin to a line meets it at the point $(-2, 9)$. Find the equation of the line.

Vedclass Test Series

Exam Paper Generator

Online Exam Module