Answer:
Question 1:
The coordinates are A(-2,9) and B(4,8)
(i) Slope = [tex]\frac{rise}{run}[/tex]
Slope = [tex]\frac{y2-y1}{x2-x1}[/tex]
Slope = [tex]\frac{8-9}{4+2}[/tex]
Slope = [tex]\frac{-1}{6}[/tex]
(ii) M(x,y) = [tex](\frac{x1+x2}{2} , \frac{y1+y2}{2} )[/tex]
=> M(x,y) = [tex](\frac{-2+4}{2}, \frac{9+8}{2})[/tex]
=> M(x,y) = [tex](\frac{2}{2} , \frac{17}{2} )[/tex]
=> M(x+y) = (1 , 8.5)
(iii) Distance Formula = [tex]\sqrt{(x2-x1)^2+(y2-y1)^2}[/tex]
D = [tex]\sqrt{(4+2)^2+(8-9)^2}[/tex]
D = [tex]\sqrt{(6)^2+(-1)^2}[/tex]
D = [tex]\sqrt{36+1}[/tex]
D = [tex]\sqrt{37 }[/tex] units
Question 2:
(i) The given equation is:
=> 3y = 6x+9
=> 3y = 3(2x+3)
Dividing both sides by 3
=> y = 2x+3
Where Slope = m = 2 and y-intercept = b = 3
Parallel lines have equal slopes
=> So, Slope of line "t" = 2
(ii) Line 3y = 6x+9 has a slope of 2
=> Perpendicular lines have a slope of negative reciprocal such that multiplying their slopes will give -1
=> Gradient of line "r" = -1/2
