The equation of the line bisecting the line s | vedclass

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

Question

(A) The slope of the line segment joining $(a, b)$ and $(a', b')$ is $m_1 = \frac{b' - b}{a' - a}$.Since the required line is the perpendicular bisect

Vedclass · Accepted Answer

$2(a - a')x + 2(b - b')y = a^2 + b^2 - a'^2 - b'^2$

Vedclass · Answer

(A) The slope of the line segment joining $(a, b)$ and $(a', b')$ is $m_1 = \frac{b' - b}{a' - a}$.Since the required line is the perpendicular bisector,its slope $m$ is $-\frac{1}{m_1} = \frac{a - a'}{b' - b} = \frac{a' - a}{b - b'}$.The midpoint of the segment is $\left( \frac{a + a'}{2}, \frac{b + b'}{2} \right)$.The equation of the line is $y - \frac{b + b'}{2} = \frac{a' - a}{b - b'} \left( x - \frac{a + a'}{2} \right)$.Multiplying by $2(b - b')$,we get $2(b - b')y - (b^2 - b'^2) = 2(a' - a)x - (a'^2 - a^2)$.Rearranging terms: $2(a - a')x + 2(b - b')y = a^2 + b^2 - a'^2 - b'^2$.

The equation of the line bisecting the line segment joining the points $(a, b)$ and $(a', b')$ at a right angle is

Similar Questions

The equation of the line perpendicular to the line $ax + by + c = 0$ and passing through $(a, b)$ is:

Sum of values of $k$ for which the image of the point $(\lambda^2 + 1, \lambda)$ with respect to the line $y = -3x + 6k$ is the point $(\lambda, \lambda - 1)$ is:

The locus of the image of a variable point $P(\alpha, 2 \alpha-1)$ with respect to the line $3 x-2 y+4=0$ is

The point $(2,3)$ is first reflected in the straight line $y=x$ and then translated through a distance of $2$ units along the positive direction of the $X$-axis. The coordinates of the transformed point are

$A$ person standing at the junction (crossing) of $2$ straight paths represented by the equations $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$,wants to reach the path whose equation is $6x - 7y + 8 = 0$ in the least time. The equation of the path he should follow is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module