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Two points A (-2, 9) and B (4, 8) lie on a line l. (i) Find the slope of the line l. (2)marks (ii) Find the coordinates of the midpoint of the points A and B (iii) Find the distance between points A and B. (i). A line “t” is parallel to 3y = 6x + 9. Find the slope of this line “t”. (ii) Another line “r” is perpendicular to the line 3y = 6x + 9. Find the gradient of the line “r”.

User BlueIceDJ
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Answer:

Question 1:

The coordinates are A(-2,9) and B(4,8)

(i) Slope =
(rise)/(run)

Slope =
(y2-y1)/(x2-x1)

Slope =
(8-9)/(4+2)

Slope =
(-1)/(6)

(ii) M(x,y) =
((x1+x2)/(2) , (y1+y2)/(2) )

=> M(x,y) =
((-2+4)/(2), (9+8)/(2))

=> M(x,y) =
((2)/(2) , (17)/(2) )

=> M(x+y) = (1 , 8.5)

(iii) Distance Formula =
√((x2-x1)^2+(y2-y1)^2)

D =
√((4+2)^2+(8-9)^2)

D =
√((6)^2+(-1)^2)

D =
√(36+1)

D =
√(37 ) 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

User Bvs
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