Question

suppose the point P=(1,2) lies on line L. Suppose that the angle between the line and the vector N = <3,4> is 90° (whenever this happens wa say vector N is normal to the line). Let Q = (x,y) be another point on the line L. Use the fact that N is orthogonal to PQ to obtain an equation of the line L.

Answer 1

Answer on Question #55094 – Math – Analytic Geometry

Suppose the point $P = (1,2)$ lies on line L. Suppose that the angle between the line and the vector $N = <3,4>$ is $90{}^\circ$ (whenever this happens we say vector N is normal to the line). Let $Q = (x,y)$ be another point on the line L. Use the fact that N is orthogonal to PQ to obtain an equation of the line L.

Solution

Vector $\overrightarrow{PQ} = \langle x - x_0, y - y_0 \rangle = \langle x - 1, y - 2 \rangle$ lies on the line L.

If $\vec{N}$ is orthogonal to $\overrightarrow{PQ}$, then the general form of an equation of the line $L$ is the following:

$$n1(x - x_0) + n2(y - y_0) = 0,$$

where $\vec{N} = \langle n1, n2 \rangle$ is a normal vector and $P(x_0, y_0)$ is a point, which lies on the line L.

So in this case we have $\vec{N} = \langle 3,4 \rangle$ and $P(1,2)$, that is, $n1 = 3$, $n2 = 4$, $x_0 = 1$, $y_0 = 2$.

After substitution we shall obtain

$$3(x - 1) + 4(y - 2) = 0 \Rightarrow 3x - 3 + 4y - 8 = 0 \Rightarrow 3x + 4y - 11 = 0.$$

Answer: $3x + 4y - 11 = 0$

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