Using one point and an angle to find the equation of a line? (A Level Maths) A Level Pure maths question
Using one point and an angle to find the equation of a line, gradient unknown? I understand how to use one point and a gradient to find an equation, but I'm confused with how to use an angle 1)b)ii)
Link to
question 1)b)ii) - "The angle BAD is a right angle. Find an equation of the line AD, giving your answer
in the form mx þ ny þ p ¼ 0 , where m, n and p are integers."
 A: Just asking, are you from UK? I do not understand gradient, so I'm going to assume that is is the slope: for a line containing points $(x_1, y_1)$ and $(x_2, y_2),$ the slope $m$ is $\frac{y_1 - y_2}{x_1 - x_2}.$ If that's what you're talking about, good. If it isn't, comment below.
I will assume that trapezium is a quadrilateral with exactly one pair of parallel sides.
Now, for the solution:
$\overline{AD} \perp \overline{AB},$ so if we can write one of the lines in the form $y = mx + b$ where $m$ is the slope and $b$ is the y-intercept, we can know the slope of the other ($-\tfrac{1}{m},$ because perpendicular lines have slopes that are negative reciprocals of each other). We proceed to rearrange the equation of $\overline{AB}.$
\begin{align*}
2x + 3y &= 14 \\
3y &= -2x + 14 \\
y &= -\dfrac{2}{3}x + \dfrac{14}{3}
\end{align*}
So we have the slope of $\overline{AD}$ is $\frac{3}{2}.$ Now we can write the equation of $\overline{AD}$ in the form $y - y_1 = m(x - x_1),$ where $(x_1, y_1)$ is a point and $m$ is the slope. ($D = (3, 7).$) We can then rearrange.
\begin{align*}
y - 7 &= \dfrac{3}{2}(x - 3) \\
2(y - 7) &= 3(x - 3) \\
2y - 14 &= 3x - 9 \\
2y - 3x - 5 &= 0~\square
\end{align*}
A better way of the answer is $\mathbf{3x - 2y + 5 = 0}.$
Now, if this is from a school math book, don't tell your teacher. ;)
Darn, sniped by John Doe.
A: AB has gradient $-2/3$ and AD is perpendicular to this, so AD has gradient $3/2$, since the gradients of two perpendicular lines multiply to give $-1$. You then know how to do the rest.
A: This is very easy to do if you use the normal form of the equation of a line $\mathbf n\cdot(\mathbf x-\mathbf p)=0$. In this equation, $\mathbf n$ is a vector normal (perpendicular) to the line, and $\mathbf p$ is any fixed point on the line.  
From the equation of line $AB$ we can read that the vector $(2,3)$ is perpendicular to this line, and so, since $\angle{BAD}$ is a right angle, is parallel to $AD$. We can get a vector perpendicular to it by swapping its coordinates and negating one of them, so an equation for the line $AD$ is $$(3,-2)\cdot(x-3,y-7)=0,$$ which becomes $3x-2y+5=0$ upon simplification.
