Question

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

Explanation:

Step1: Find the equation of line \( \ell \)

First, find the slope \( m \) of line \( \ell \) using the two points \( (0, -3) \) and \( (7, 4) \). The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Substituting the values, we get \( m=\frac{4 - (-3)}{7 - 0}=\frac{7}{7} = 1 \). Using the point - slope form \( y - y_1=m(x - x_1) \) with the point \( (0,-3) \) (where \( x_1 = 0,y_1=-3 \) and \( m = 1 \)), the equation of the line is \( y-(-3)=1\times(x - 0) \), which simplifies to \( y=x - 3 \), or in standard form \( x - y-3 = 0 \).

Step2: Use the distance formula from a point to a line

The formula for the distance \( d \) from a point \( (x_0,y_0) \) to a line \( Ax+By + C = 0 \) is \( d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^2 + B^2}} \). For the line \( x - y-3 = 0 \) (so \( A = 1,B=-1,C=-3 \)) and the point \( P(4,3) \) (so \( x_0 = 4,y_0 = 3 \)), we substitute into the formula:
\( d=\frac{\vert1\times4+(-1)\times3-3\vert}{\sqrt{1^2+(-1)^2}}=\frac{\vert4 - 3-3\vert}{\sqrt{1 + 1}}=\frac{\vert-2\vert}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\approx1.414 \)

Answer:

C) $\sqrt{2}$ or about 1.414 units