Question

find the distance between point p and line (ell). line (ell) contains points ((0, - 3)) and ((7,4)). point p has coordinates ((4,3)). a) (sqrt{3}) or about 1.73 units b) (sqrt{10}) or about 3.16 units c) (sqrt{2}) or about 1.41 units d) 2 units

Explanation:

Step1: Find the slope of line $\ell$

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For points $(0,-3)$ and $(7,4)$ on line $\ell$, we have $m=\frac{4-(-3)}{7 - 0}=\frac{7}{7}=1$.

Step2: Find the equation of line $\ell$

Using the point - slope form $y - y_1=m(x - x_1)$ with the point $(0,-3)$ and $m = 1$, we get $y-(-3)=1(x - 0)$, which simplifies to $y=x - 3$ or $x-y-3=0$.

Step3: Use the distance formula from a point to a line

The distance $d$ from a point $(x_0,y_0)$ to a line $Ax+By + C=0$ is given by $d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^2 + B^2}}$. For point $P(4,3)$ and line $x-y-3=0$ (where $A = 1$, $B=-1$, $C=-3$, $x_0 = 4$, $y_0=3$), we have $d=\frac{\vert1\times4+(-1)\times3-3\vert}{\sqrt{1^2+(-1)^2}}=\frac{\vert4 - 3-3\vert}{\sqrt{2}}=\frac{\vert-2\vert}{\sqrt{2}}=\sqrt{2}\approx1.41$ units.

Answer:

C) $\sqrt{2}$ or about $1.41$ units