# Please help me with the question below.17. In the xy-plane, line l passes through the origin and is perpendicular to the line 4x + y = k, where k is a constant. If the two lines intersect at...

Please help me with the question below.

17. In the *xy-*plane, line *l* passes through the origin and is perpendicular to the line 4*x* + *y* = *k*, where *k* is a constant. If the two lines intersect at the point (*t, t* + 1), what is the value of *t*?

(A) -4/3

(B) -5/4

(C) 3/4

(D) 5/4

(E) 4/3

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The line l passes through the origin and is perpendicular to the line 4x + y = k.

4x + y = k ...(1)

=> y = k - 4x

This is in the form y = mx + c, where m is the slope of the line. The slope of 4x + y = k is -4. As the product of the slope of two perpendicular lines is -1, line perpendicular to this has a slope -1/(-4) = 1/4

The line l has a slope 1/4 and passes through (0, 0). The equation of the line is y/x = 1/4

=> 4y = x ...(2)

Solving (1) and (2) gives the point of intersection of the two lines.

Substitute x = 4y in (1)

=> 4(4y) + y = k

=> 16y + y = k

=> 17y = k

=> y = k/17

x = 4k/17

The point of intersection of the lines is (t, t+1)

t = 4k/17 and t + 1 = k/17

t + 1 = k/17

=> t = (k - 17)/17

4k = k - 17

=> 3k = -17

=> k = -17/3

t = (-17/3 - 17)/17 = -1/3 - 1 = -4/3

**The correct option is A, t = -4/3**

if line l is perpendicular to 4x+y=k and goes through the origin, then the equation of line l is:

slope of line l = negative reciprocal of the slope of 4x+y=k

slope of (4x+y=k) = -4

slope of line l = 1/4

equation of line l is y = x/4 or 4y=x

4(4y)+y=k substitute 4y for x

16y+y=k

17y=k

y=k/17 ---> t+1=k/17 substitute t+1 for y and t for x

x=4k/17---> t=4k/17 <-------,

4k/17 +1 = k/17 | substitute 4k/17 for t

4k + 17 = k | and solve for k

3k = -17 |

k=-17/3 ----------------------' substitute back

t = 4(-17/3)/17 the 17 divides and cancels out

t = -4/3

the answer is A