I have a question I'm working on. I don't really understand it.
Find the value of $k$ for which the graphs of $2y + x + 3 = 0$ and $2y + kx + 2 = 0$ intersect at right angles.
I'm used to line equations with the formula $y = mx + b$. I don't really understand this. I also don't know exactly what is meant by "intersect at right angles." Can anyone provide some more insight to this? Thanks!
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$\begingroup$If two lines intersect at right angles, they'll be perpendicular, and the slopes of two perpendicular lines are opposite reciprocals. That is, if a certain line has a slope of $m$, then a line that's perpendicular to that first line will have a slope of $-1/m$.
In order to find the value of $k$ such that those two lines are perpendicular, you'll first need to solve each equation for $y$ (that is, isolate $y$ on one side of each equation by moving all the other terms to the other side), and once you've done that, it shouldn't be too hard to figure out what value of $k$ will make the slope of one line the opposite reciprocal of the slope of the other line.
Here's an example of two perpendicular lines. In this case, the equation of the red line is $y=-2x+4$, and the equation of the blue line is $y=\frac{1}{2}x+1$. Notice that $\frac{-1}{\text{either slope}}$ gives you the other slope.
The condition for two lines to intersect at right angles is that the product of their gradients (slopes) is $-1$.
Put both equations in the standard form $y = mx+c$ where $m$ is the gradient.
$2y+x+3 = 0 \implies y =-\frac 12x - \frac 32$
Gradient is $-\frac 12$. Hence other line must have gradient of $\frac{-1}{-\frac 12}=2$.
$2y+kx+2=0 \implies y=-\frac k2 x - 1$
Gradient is $-\frac k2$.
So $-\frac k2 = 2 \implies k=-4$.
Note that only the gradient is important here. The other term ($y$-intercept) is irrelevant.
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