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In Exercises pointslopegivenlinefirst - pointslopegivenlinelast, find both the point-slope form and the slope-intercept form of the line with the given slope which passes
through the given point.
\(m = 3, \;\; P(3, -1)\)
The point slope form is ...
\(y+1=3(x-3)\)
The slope-intercept form is
\(y = \answer {3}x + \answer {-10}\)
\(m = -2, \;\; P(-5, 8)\)
The point slope form is ...
\(y-8 = -2(x+5)\)
The slope-intercept form is
\(y = \answer {-2}x + \answer {-2}\)
\(m = -1, \;\; P(-7, -1)\)
The point slope form is ...
\(y + 1 = -(x+7)\)
The slope-intercept form is
\(y = \answer {-1}x + \answer {-8}\)
\(m = \frac {2}{3}, \;\; P(-2, 1)\)
The point slope form is ...
\(y - 1 = \frac {2}{3} (x+2)\)
The slope-intercept form is
\(y = \answer {\frac {2}{3}} x + \answer {\frac {7}{3}}\)
Find all of the points on the line \(y=2x+1\) which are \(4\) units from the point \((-1,3)\).
\((-1,-1)\) and \(\left (\frac {11}{5}, \frac {27}{5}\right )\)
In Exercises parallelfirst - parallellast, you are given a line and a point which is not on that line. Find the line parallel to the given line which passes
through the given point.
\(y = 3x + 2, \; P(0, 0)\)
\(y = \answer {3}x\)
\(y = -6x + 5, \; P(3, 2)\)
\(y = \answer {-6}x + \answer {20}\)
\(y = \frac {2}{3} x - 7, \; P(6, 0)\)
\(y = \answer {\frac {2}{3}} x + \answer {-4}\)
\(y = \frac {4-x}{3}, \; P(1, -1)\)
\(y = \answer {-\frac {1}{3}} x + \answer {-\frac {2}{3}}\)
\(y = 6, \; P(3, -2)\)
\(y= \answer {-2}\)
\(x=1, \; P(-5,0)\)
\(x= \answer {-5}\)
In Exercises perpendlinefirst - perpendlinelast, you are given a line and a point which is not on that line. Find the line perpendicular to the given line which
passes through the given point.
We shall now prove that \(y = m_1x + b_1\) is perpendicular to \(y = m_2x + b_2\) if and only if \(m_1 \cdot m_2 = -1\). To make our lives easier we shall assume that \(m_1 > 0\) and \(m_2 < 0\). We can also
“move” the lines so that their point of intersection is the origin without messing things up, so we’ll assume \(b_1 = b_2 = 0.\) (Take a moment
with your classmates to discuss why this is okay.) Graphing the lines and plotting the points \(O(0, 0)\;\), \(P(1, m_1)\;\) and \(Q(1, m_2)\) gives us the following set
up.
The line \(y = m_1x\) will be perpendicular to the line \(y = m_2x\) if and only if \(\bigtriangleup OPQ\) is a right triangle. Let \(d_1\) be the distance from \(O\) to \(P\), let \(d_2\) be the distance
from \(O\) to \(Q\) and let \(d_3\) be the distance from \(P\) to \(Q\). Use the Pythagorean Theorem to show that \(\bigtriangleup OPQ\) is a right triangle if and only if \(m_1 \cdot m_2 = -1\) by
showing \(d_1^{2} + d_2^{2} = d_3^2\) if and only if \(m_1 \cdot m_2 = -1\).
![[Picture]](ExercisesforAppLines.online-271e3969e453bda5f27e89294dd5183b.svg)