Find the equation of a 2D line which...
A) Passes through the point A(0,-2) and is perpendicular to the vector v=(-2,-5)
B)Passes through the points A(2,-3) and B(3,-4)
C)Passes through the point A (2,-4) and is parallel to the line y=-2x-5
I tried to do these questions and so far I only understood how to do B) which is y=-x-1
On
For A), the most direct way is to use the vector form of the equation: if $(x_A,y_A)$ are the coordinates of the point $A$, and $\vec n$ is a normal vector, the equation is $$\overrightarrow{AM}\cdot \vec n=0,\quad\text{i.e.}\quad 2x+5(y+2)=0.$$
For B), if the given poibts have different abscissae, there's a formula, which is a translation of the colinearity of the vectors $\overrightarrow{AM}$ and $\overrightarrow{AB}$: $$\frac{y -y_A}{x-x_A}=\frac{y_B -y_A}{x_B-x_A}\iff y=\frac{y_B -y_A}{x_B-x_A}(x-x_A)+y_A.$$
For C), use the formula for the eqaution of a line passing through $A$, with prescribed slope $m$: $$y=m(x-x_A)+y_A.$$
For A and C, I would use the point-slope form of a line. It states that a line through $(u, v)$ with slope $m$ has the form $y-v = (x-u)m $.
Note that this is true when $x=u$ and $y=v$, so it passes through the point. Also note that the equation can be written $\dfrac{y-v}{x-u} =m $, so its slope is $m$.
The vector through $(-2, -5)$ (and, I assume, $(0, 0)$) has slope $\dfrac{-5}{-2} =\dfrac{5}{2} $, so the slope of the normal to it has a value which is its negative reciprocal which is $\dfrac{-2}{5}$.
For C, the slope of the line $y=-2x-5$ is $-2$, and it passes through $ (2,-4)$.
You should now be able to get your answers.