Equation of a straight line help me to solve this
1. the vertices of a rectangle are A(0,0), B(-8,0), C(-8,-6) and D(0.-6). write equations for the lines witch form its sides and diagonals
x = -8, 3x - 4y = 0
x = 0, x - 4y = 0
2. the vertices of a rectangle are A(2,6), B(8,-4) and C(-3,-5). two lines are drawn through A. one parallel to and the other perpendicular to BC. find the equation of each of these lines
x - 11y + 64 = 0
11x + y - 28 = 0
3. write the equation of the line that is the perpendicular bisector of the given segment AB. A(-2,4), B(4,0)
3x - 2y + 1 = 0
Diagonal AC :`y=6/8 x=3/4x`
Diagonal BD: slope`=(-6-0)/(0-(-8))=-6/8=-3/4`
Slope point equation
Equation of BD: `y=-(3/4) x+8`
3.Let M be the slope of AB.
for a line to be perpenducilar to this, it's slope (let's say K) must be 3/2, since K*M =-1 ALWAYS for perpendicular lines!
The midpont of AB is C(1,2) found with midpoint formula
[(x1+x2)/2 , (y1+y2)/2)
The line that passes through C(1,2) with a slope K=3/2 is:
2y-3x = 4-3
this was done using the formula:
y-y1=K(x-x1) where x1 and y1 are the coordinates of the point through which it should pass and K is the slope.