# In the xy-plane, the line L passes through the origin and is perpendicular to the line 4x+y = k, where k is?

the constant. If the two lines intersect at the point (t, t+1), what is the value of t?

### 4 Answers

- Anonymous1 decade agoFavorite Answer
In Ax + By = C, m = -A / B.

The slope of the line 4x + y = k is -4 / 1 = -4.

The slope of the line perpendicular (passing through the origin) is its negative reciprocal, or 1/4.

The equation of this line,

y = mx + b = (1/4)x + 0.

y = (1/4)x, or x - 4y = 0.

The lines intersect at (t, t + 1), and this is the solution to the system of equations for both lines.

In the second line,

x - 4y = 0

(t) - 4(t + 1) = 0

-3t - 4 = 0

-3t = 4

t = -4/3

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- cattbarfLv 71 decade ago
'where k is?" are we doing math in eubonics these days?

If L passes thru the origin, its intercept is 0. If it perpendicular to a line with slope -4, L has a slope of 1/4. So we have y= x/4 and y = k - 4x. Where they intersect, x=t and y=t+1. Then

t+1 = t/4 , and t = -3/4 , t+1 = 1/4 . Substituting, k= -2&3/4.

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- clinkscalesLv 44 years ago
First rewrite the perpendicular line in slope intercept form: y = -4x + ok So the line L's slope must be the unfavourable reciprocal of -4 and because it passes via the inspiration this is y-intercept (b) = 0 So the equation of line L is y = (a million/4)x to locate t, plug (t, t+a million) in to the equation y = (a million/4)x t+a million = (a million/4)t a million = (-3/4)t -4/3 = t

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- NorthstarLv 71 decade ago
The equation of the given line is 4x + y = k.

The given line passes thru the point (t, t+1), where t is a constant.

Line L passes thru the origin and is perpendicular to the given line. The slope of this line is the negative reciprocal of the given line. To write a line with that slope, swap the coefficients of x and y and reverse the sign of one of them. The equation of line L is:

x - 4y = 0

Plug in the point (t, t+1) for (x,y) on line L.

t - 4(t + 1) = 0

t - 4t - 4 = 0

-3t = 4

t = -4/3

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