If (-2, 6) is the image of the point (4, 2) w | vedclass

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

Question

(A) Let the point $P = (-2, 6)$ and $Q = (4, 2)$. The line $L = 0$ is the perpendicular bisector of the segment $PQ$.The midpoint of $PQ$ is $\left( \

Vedclass · Accepted Answer

$3x - 2y + 5 = 0$

Vedclass · Answer

(A) Let the point $P = (-2, 6)$ and $Q = (4, 2)$. The line $L = 0$ is the perpendicular bisector of the segment $PQ$.The midpoint of $PQ$ is $\left( \frac{-2 + 4}{2}, \frac{6 + 2}{2} \right) = (1, 4)$.The slope of $PQ$ is $m_{PQ} = \frac{2 - 6}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}$.Since line $L$ is perpendicular to $PQ$,its slope $m_L$ is the negative reciprocal of $m_{PQ}$,so $m_L = \frac{3}{2}$.The equation of line $L$ passing through $(1, 4)$ with slope $\frac{3}{2}$ is given by:$y - 4 = \frac{3}{2}(x - 1)$$2(y - 4) = 3(x - 1)$$2y - 8 = 3x - 3$$3x - 2y + 5 = 0$.

If $(-2, 6)$ is the image of the point $(4, 2)$ with respect to line $L = 0$,then $L =$

Similar Questions

Which one of the following statements is not correct?

Particles of masses $m, 2m, 3m, \dots, nm$ grams are placed on the same line at distances $l, 2l, 3l, \dots, nl$ cm from a fixed point. The distance of the centre of mass of the particles from the fixed point in centimetres is:

The wavelengths of two sound notes in air are $\frac{40}{195} ~m$ and $\frac{40}{193} ~m$. Each note produces $9$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m/s$ is

Which of the following is an oxidizing agent?

The internal energy change in a system that has absorbed $2 \, kcal$ of heat and done $500 \, J$ of work is ........... $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module