Equations of straight lines

用戶225241

2008-05-26 6:17 am

AB is a st. line of slope -3/4 passing through through the point(3,0)
a. Find the equations of AB
b. Find the coordinates of the points P and Q on the line x-y=0 with a
perpendicular distance 1 unit from AB
c. Find the equations of the two st. lines passing through Pand Q
respectively and equidistant from the line AB
d. If AB =3, find the area of all the triangles formed by A, B and the third vertex on either one of the two lines in (c)

回答 (1)

sha wan

2008-05-26 8:07 am

✔ 最佳答案

a. equations of AB
-3/4 = (y-0)/(x-3)
-3x + 9 = 4y
0 = 4y + 3x - 9
b.
let co-ordinates of P be (p, p)
Since the distance is 1, and it may be + or -
+-1 = (4p + 3p - 9) / √(4^2 + 3^2)
+-5 = (4p + 3p - 9)
9 +- 5 = 7p
4 or 14 = 7p
4/7 or 2 =p
The co-ordinates of P and Q are (2, 2) and (4/7, 4/7)
c.
As they are equidistant from the line AB, the lines are parallel to AB
and the slope is -3/4
the line passes through (2, 2) is
-3/4 = (y-2)/(x-2)
-3x + 6 = 4y-8
0 = 4y+3x -14

the line passes through (4/7, 4/7) is
-3/4 = (y-4/7)/(x-4/7)
-3/4 = (7y-4)/(7x-4)
-21x +16 = 28y-16
0 = 28y+21x -32
d. whatever the point is , the height is 1
area of triangle = 3(1)/2 = 1.5

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