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a line passes through the point of intersection of the lines $y = -\frac12x - 6 $ and $y = 2x + 4$. determine the equation of the line if it has a slope of $\frac12$.

I'm completely lost, how should i answer this? I know that the line that has a slope of $\frac12$ is intersecting through $y = -\frac12x - 6$ and $y = 2x + 4$, which means that the points to these intersecting lines are the same to the line i'm trying to find the equation for, so how should i figure out those points? (unless my method is wrong)

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closed as off-topic by 5xum, user8795, Namaste, Henrik, Glorfindel Aug 16 '17 at 20:39

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OK, you are looking for a line, so you are looking for $k$ and $n$ in the equation $y=kx+n$. Basically, you want to calculate $k$ and $n$.

Hint:

  1. What is the slope of the line $y=kx+n$?
  2. What is the intersection of the lines $y=-\frac12 x - 6$ and $y=2x+4$?
  3. If I know that the point $(x_0,y_0)$ is on the line $y=kx+n$, what equation do $x_0$ and $y_0$ satisfy?
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  • $\begingroup$ 1) 1/2, 2) L1 (0, -6) L2 (0,4), so do i use one of these points to satisfy the equation and solve for n? $\endgroup$ – Jenny B Aug 2 '17 at 6:45
  • $\begingroup$ @ineedgr10mathhelp wrong and wrong. For 1) I am asking in general, if I give you an equation $y=kx+n$, what is the slope of this line? Like, what is the slope of the line $y=2x+5$? And what is the slope of $y=x+1$? For 2) no. Those are the intersections of the two lines with the $y$ axis, but not the intersection of the two lines among themselves. You want to find a point that is on both lines! $\endgroup$ – 5xum Aug 2 '17 at 6:51
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Based on the hints @5xum gave you, you should see a connection between y=kx+n and y=mx+b (which is what most teachers use).

  1. The slope of y=kx+n is k. The question gives that to you.

  2. You can find the intersection of the two lines algebraically. Since y=y, then -1/2 x - 6 = 2x + 4. Now solve for x. I personally prefer to graph the lines on Desmos.

  3. When you find the point, plug in what you know to y=kx+n. So far, you know k (slope) and x&y (the point of intersection). Remember to multiply k times x. You are left with n as the unknown so solve for n. (Get the n by itself on one side of the equation and a number on the other side.)

  4. For your final equation, set up y=kx+n and fill in the k (slope) and n (y-intercept of the new line).

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  • $\begingroup$ Most teachers where I'm from use $y=kx+n$, not $y=mx + b$... $\endgroup$ – 5xum Aug 2 '17 at 6:57
  • $\begingroup$ I suppose it could be a regional difference. I have occasionally seen y=ax+b $\endgroup$ – MrsSnider Aug 2 '17 at 7:00
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The intersection point of the lines $y = -\frac12x-6$ and $y = 2x+4$ is a point where the value of $y$ for both the lines is same. So, $$ -\frac12x-6 = 2x+4 $$ $$-\frac52x = 10 $$ $$x = -4 $$ So the point of intersection is at $x=-4$ and $y = -\frac12x-6 = -\frac12(-4) - 6 = -4$. The point of intersection is $(-4, -4)$. Now we know that the unknown line with slope $\frac12$ passes through the point $(-4, -4)$.

I hope this helps.

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Any line passing through the intersection of the given lines is

$$ \lambda( -x/2-6) +(1-\lambda)(2 x +4) = x/2 + C $$

Put $x=0$ in the above equation,it gives

$-10 \lambda +4 = C \tag1$

Collecting $x$ term coefficients and equating

$$ -\lambda/2 + 2 - 2\lambda = \frac12 ,\, \lambda =\frac35 \tag2 $$

Eliminate $\lambda$ from (1), (2)

$$C= -2, \tag3$$

So required straight line has equation

$$ y = x/2 -2.$$

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