Equations of straight lines

2010-11-11 5:46 am
1. Find the equation of straight lines satisfying the following conditions:
The product of x-intercept and the y-intercept is 20, and parallel to the line
5x+2y-6=0

2. Find the equation of line(s) such that perpendicular to L1: x-y=0, and the
area of the triangle enclosed by the line and the axes is 7/1/5 .

回答 (1)

2010-11-11 8:05 am
✔ 最佳答案
1. Find the equation of straight lines satisfying the following conditions: The product of x-intercept and the y-intercept is 20, and parallel to the line 5x+2y-6=0

Since the required line is parallel to the line 5x + 2y - 6 = 0
Let 5x + 2y + c = 0 be the required line.

When x = 0, y = -c/2
Hence, y-intercept of the required line = -c/2

When y = 0, x = -c/5
Hence, x-intercept of the required line = -c/5

The product of the x-intercept and the y-intercept:
(-c/2)(-c/5) = 20
c² = 200
c = 10√2 or c = -10√2

The required line is:
5x + 2y + 10√2 = 0 or 5x + 2y - 10√2 = 0


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2. Find the equation of line(s) such that perpendicular to L1: x - y = 0, and the area of the triangle enclosed by the line and the axes is 7/1/5 .

Slope of L1 = -1/(-1) = 1
Slope of the required line = -1/1 = -1

Let x + y + c = 0 be the required line.

When x = 0, y = -c
y-intercept = -c

When y = 0, x = -c
x-intercept = -c

Area of the triangle enclosed by the line and the axes:
c²/2 = (7 and 1/5)
c²/2 = 36/5
c² = 360/25
c = (6/5)√10

The required line is:
x + y + (6/5)√10 = 0 or x + y + (6/5)√10 = 0
參考: 土扁


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