# Equations of straight lines

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 .

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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

資料來源： 土扁
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