Question

Can anybody help me with this

Answer 1

A = (4, 5)   B = (-2, 1)

Midpoint of A and B

$C=(X_M, Y_M) = \bigg(\dfrac{X_A+X_B}{2}, \dfrac{Y_A+Y_B}{2}\bigg)\\\\. \qquad \qquad \qquad =\bigg(\dfrac{4-2}{2},\dfrac{5+1}{2}\bigg)\\\\. \qquad \qquad \qquad =\bigg(\dfrac{2}{2},\dfrac{6}{2}\bigg)\\\\. \qquad \qquad \qquad =(1, 3)$

Distance from A to B

$d_{AB}=\sqrt{(X_B-X_A)^2+(Y_B-Y_A)^2}\\\\.\qquad =\sqrt{(-2-4)^2+(1-5)^2}\\\\.\qquad =\sqrt{(-6)^2+(-4)^2}\\\\.\qquad =\sqrt{36+16}\\\\.\qquad =\sqrt{52}\\\\.\qquad =7.2$

Equation of line through AB

$m_{AB}=\dfrac{Y_B-Y_A}{X_B-X_A}\\\\.\qquad =\dfrac{1-5}{-2-4}\\\\.\qquad =\dfrac{-4}{-6}\\\\.\qquad =\dfrac{2}{3}$

$Y-Y_A=m_{AB}(X-X_A)\\\\Y-5=\dfrac{2}{3}(X-4)\\\\Y-5=\dfrac{2}{3}X-\dfrac{8}{3}\\\\Y=\dfrac{2}{3}X+\dfrac{7}{3}$

Line parallel to AB (same slope as AB) through point (3, -5)

$Y-Y_A=m_{AB}(X-X_A)\\\\Y+5=\dfrac{2}{3}(X-3)\\\\Y+5=\dfrac{2}{3}X-2\\\\Y=\dfrac{2}{3}X-7$

AB is perpendicular to A'B' so slopes are opposite reciprocals

A' = (3, 0)

B' = (-1, 6)

C = (1, 3)

C' = (7, 3)

D = (5, 3)

D' = (5, 1)