How to Find Distance between a Point $(-4,2)$ and a Line having equation $3x−y−10=0$

Question

Find Distance between a Point $(-4,2)$ and a Line having equation $3x−y−10=0$

My attempt:

I used a graphing calculator to graph the equation $3x−y−10=0$ and the point $(-4,2)$. I found the x-intercept and y-intercept of the equation as $\dfrac{10}{3}$, and $-10$ respectively.

I used the Midpoint Formula to get the midpoint of $AB$;

$$ \left(x_{m}, y_{m}\right)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) $$ $$= \left(\dfrac{\dfrac{10}{3} + 0}{2}, \; \dfrac{0 + -10}{2}\right) = \left(\dfrac{5}{3}, \; -5\right)$$

Now, Can I use the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ to find the distance between the point and the midpoint of the line? By distance formula, $d=9$. Is my approach correct?

But the solution given in my textbook uses some other distance formula, $d=\left| \dfrac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|$. Can you please explain to me how we got this formula and how to solve this problem with this formula?

Answer 1

By the distance formula, I found the length of $AB$ as $\dfrac{10}{3} \sqrt{10}$ units. Now, suppose that $PQ$ is the perpendicular drawn from $P$ to $\overleftrightarrow{l}$. Now, the area of a triangle is given as:

$$Area =\dfrac{1}{2}× Base × Height$$

So, using this result, we can write: \begin{align*} &\hphantom{{}\implies{}} \text{area}\,\triangle PAB = \frac{1}{2} \times AB \times PQ \\ &\implies \text{area}\,\triangle PAB = \frac{1}{2} \times \frac{10\sqrt{10}}{3} \times PQ \\ &\implies PQ = \frac{3}{5\sqrt{10}} \text{area}\,\triangle PAB \end{align*}

Consider the above △PAB again; Now, the area of a triangle having vertices at $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is given as $$ Area = \frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right| $$

So, using the above result, we can say that Area △PAB= 40 sq. units. Now, substituting the value of ‘Area △PAB’, we’ll get $PQ=\dfrac{24}{\sqrt{10}}$.

Therefore, the distance between the point $P$ and the line $\overleftrightarrow{l}$ is $\dfrac{24}{\sqrt{10}}$ units.

Answer 2

Forget about formula. In the following draw, the distance between $P$ and the line passing through $A$ and $B$ is $\|\vec{PH}\|$. Now, $$\vec{PH}=\vec{AH}-\vec{AP}.$$

You know $\vec{AP}$. Now, $\vec{AH}$ is the orthogonal projection of $\vec{AP}$ on $\vec{AB}$, and this is $$\left<\vec{AP},\frac{\vec{AB}}{\|\vec{AB}\|}\right>\frac{\vec{AB}}{\|\vec{AB}\|}.$$

You have all element to conclude.

PS : Given your picture, to simplify calculation, you should replace my $A$ by your $B$.

Answer 3

Your line is $$3x−y−10=0$$ Therefore we have $a=3$ and $b=-1$ and $c=-10$

Your point is $$(x_1, y_1)=(-4,2)$$

You have the distance formula, $$ d=\left| \dfrac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|$$

You should be able to find the distance.

There is no need for the midpoint of $AB$

Answer 4

You don't need to use vectors, you can use coordinate geometry.

Formulate a strategy.

1. First find a line $l_2$ perpendicular to $3x-y-10=0$ (call it line $l_1$) that also passes through $(-4,2)$ (call it point $A$).

2. Then find the intersection (call it $P$) between $l_2$ and $l_1$. This will be the foot of the perpendicular dropped from $A$ to $l_1$.

3. Then just find the distance $AP$, which will give you what you require.

For 1) find the slope of $l_2$ by taking the negative reciprocal of the slope of $l_1$. So the slope of $l_2$ is $-\frac 13$

Then use the formula $y - y_1 = m(x-x_1)$ to find the equation of $l_2$. So putting in the coordinates of $A$, $y-2 = -\frac 13(x+4) \implies y = -\frac 13x + \frac 23$

2. Now solve $l_1$ and $l_2$ simultaneously for $P$. $-\frac 13x_P + \frac 23 = 3x_P - 10 \implies x_P = \frac {16}5 \implies y_P = -\frac 25$

So $P = (\frac{16}5,-\frac 25)$

3. Finally find $AP$ using the Euclidean distance formula. $AP= \sqrt{(-4-\frac{16}5)^2 + (2+\frac 25)^2} = \sqrt{\frac{32}5}$

And so the answer is $\sqrt{\frac{32}5}$.