Question

find the distance from point p to line ab.
6.
7.

Explanation:

Step1: Analyze the distance for problem 6

Line \(AB\) is horizontal. The vertical distance from point \(P\) to line \(AB\) is calculated by counting the number of grid - units.

Step2: Calculate the distance for problem 6

The \(y\) - coordinate of line \(AB\) is \(y = 2\), and the \(y\) - coordinate of point \(P\) is \(y=-1\). The distance \(d=\vert2-(-1)\vert = 3\) units.

Step3: Analyze the distance for problem 7

Use the formula for the distance from a point \((x_0,y_0)\) to a line \(Ax + By+C = 0\). First, find the equation of line \(AB\). Points \(A(0, - 2)\) and \(B(-4,1)\). The slope \(m=\frac{1-(-2)}{-4 - 0}=-\frac{3}{4}\). Using the point - slope form \(y - y_1=m(x - x_1)\) with point \(A(0,-2)\), the equation is \(y+2=-\frac{3}{4}(x - 0)\), or \(3x+4y+8 = 0\). Point \(P(2,4)\).

Step4: Calculate the distance for problem 7

The distance formula \(d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^{2}+B^{2}}}\). Here \(A = 3\), \(B = 4\), \(C = 8\), \(x_0=2\), \(y_0 = 4\). Then \(d=\frac{\vert3\times2+4\times4 + 8\vert}{\sqrt{3^{2}+4^{2}}}=\frac{\vert6 + 16+8\vert}{5}=\frac{30}{5}=6\) units.

Answer:

1. The distance is \(3\) units.
2. The distance is \(6\) units.