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((X_A+X_B)/(2), (Y_A+Y_B)/(2)\bigg)\\\\. \qquad \qquad \qquad =\bigg((4-2)/(2),(5+1)/(2)\bigg)\\\\. \qquad \qquad \qquad =\bigg((2)/(2),(6)/(2)\bigg)\\\\. \qquad \qquad \qquad =(1, 3)`

Distance from A to B

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

Equation of line through AB

`m_(AB)=(Y_B-Y_A)/(X_B-X_A)\\\\.\qquad =(1-5)/(-2-4)\\\\.\qquad =(-4)/(-6)\\\\.\qquad =(2)/(3)`

`Y-Y_A=m_(AB)(X-X_A)\\\\Y-5=(2)/(3)(X-4)\\\\Y-5=(2)/(3)X-(8)/(3)\\\\Y=(2)/(3)X+(7)/(3)`

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

`Y-Y_A=m_(AB)(X-X_A)\\\\Y+5=(2)/(3)(X-3)\\\\Y+5=(2)/(3)X-2\\\\Y=(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)