1d斜率既問題

斜率slope = difference between y-coordinates/difference between x-coordinates
找兩點的斜率
6.R(5,6),S(4,1)
slope
= (6-1)/(5-4)
= 5

8.S(b,-4)and T(-9,-8) are two points on a straight line L,If the slope of L is 4/3,find b.

slope = difference between y-coordinates/difference between x-coordinates
4/3 = [-4-(-8)] / [b-(-9)]
4/3 = 4/ (b+9)
b+9 = 3
b =6

10.Using the slope formula,determine whether
A(-3,3),B(-1,2) and C(1,1)are collinear

slope of AB
= (3-2)/[-3-(-1)]
= -1/2
slope of AC
= (3-1)/[(-3-1)]
= -1/2
slope of AB = slope of AC, therefore A(-3,3),B(-1,2) and C(1,1)are collinear

15.P(2k,3)and Q(-3,3k)are two points on a straight line L,If the slope of L is 4/3,find k

slope of PQ = 4/3
(3-3k)/[2k-(-3)] = 4/3
9-9k = 8k + 12
17k = -3
k= -3/17

A(X1, Y1) , B(X2, Y2)

Equation: 斜率(Slope) = [A點Y軸坐標 - B點Y軸坐標] / [A點X軸坐標 - B點X軸坐標]
=[Y1-Y2] / [X1 - X2]


6.R(5,6),S(4,1)

Slope = [6-1] / [5-4] = 5/1 = 5


8.S(b,-4)and T(-9,-8) are two points on a straight
line L,If the slope of L is 4/3,find b.

Slope=4/3 = [-4-(-8)] /[b-(-9)] = 4/ [b+9]

so. 3=b+9 => b= -6

10.Using the slope formula,determine whether
A(-3,3),B(-1,2) and C(1,1)are collinear

Slope(AB) = [3-2]/ [-3-(-1)] = 1/(-2)

Slope(BC) = [2-1]/ [-1-1] = 1(-2)

Slope(CA) = [3-1]/ [-3-1] = 1/(-2)

Slope(AB)=Slope(BC)=Slope(CA), So Collinear.


15.P(2k,3)and Q(-3,3k)are two points on a straight
line L,If the slope of L is 4/3,find k

Slope=4/3 = [3-3k] /[2k-(-3)] = [3-3k] / [2k+3]

so. 4*[2k+3] = 3*[3-3k]

==> 8k+12 = 9-9k
==> 8k+9k = 9-12
==> 17k = -3
==> k = -3/17