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YOU are better than YOU think. Show yourself how:
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-/[]\- Logic chapters 1 to 5 re- appear not in sequence, as is or longer, in Volume 1A, Pattern Based Reason, Bon Appetite. Logic
Mastery Logic mastery makes the hard, easier. Logic mastery leads to better, stronger and richer comprehension. Logic mastery improves reading and writing. Logic mastery ease learning difficulties. Logic mastery gives a headstart. In sum, logic mastery will develops critical thinking, improve reading and writing, and give a firmer base for work and studies at many levels. Good luck. After logic, (a) continue reading Three Skills for Algebra, chapters 8 to 14 and do so alongside site area on solving liinear Equations ; or (b) see this calculus starter lesson and Volume 3, Why Slopes & More Math, chapters 2 to 6;
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-/[]\- What may be learnt and when depends on how skills and concepts are developed. Making the hard easier and clearer will allow earlier & richer development of skills and concepts. Try the Twiddla
Whiteboard. In principle, it allows
to people to draw and chat together online on a copy of this webpage or a clean
sheet. The chat may be via text or audio. Visit www.twiddla.com
to set up whiteboards to work with the webpage of your choice. |
Slope Product for Perpendicular LinesTheorem: If L1 and L2 are perpendicular lines, both non-vertical then the product m1m2 of their slopes m1 and m2 equals -1. That is, -1 = m1m2 Here the roles of L1 and L2 are interchangeable. Three Easy Consequences:
Proof of theorem:Since L1 and L2 are perpendicular lines. Then they will intersect at some point. We assume L2 is the upward slanting line while L1 is the downward slanting line (for travel from right to left along the line). NB. The roles of L1 and L2 can be interchanged: geometric sense suggests one of them has to be slanted downward.
Now will draw a triangle to find the rise over run of L2
The hypotenuse of the right triangle has length c while its other two sides parallel to the x ayd y axes respectively have a and b. The rise over run ratio for L2 equal b/a. So the slope of L2 is
Our next step is draw another right triangle with a vertex at the intersection, hypotenuse of length c along L1 and base parallel to the x axis. One such triangle is shown below in a copy of the first diagram above.
Here EBC forms a straight angle (180 degrees) and angle ABD is 90 degrees (or should be). Therefore we have
The some of the angles in right triangle ABC sums to 180 degrees
Comparison of these two angle sum equations gives
as left and right hand side are both equal to 180 degrees. Therefore
Therefore the remaining angles in the right triangles ABC and right triangle DEB are equal.
So corresponding angles on either side of the hypotenuse of length c in both right triangles are equal. The side-angle-side isometry postulate implies the right triangles are isometric or congruent. So corresponding lengths are equal. In particular, the vertical side in the triangle with hypotenuse on L1 has length a while its horizontal side has length b as shown in the next diagram.
Therefore the drop over run ratio or proportionality constant for L1 is a/b. The negative of this ratio gives the slope m1 of L1. So m1 = - a/b Recall the slope of L2 is
Therefore -1 = (b/a)(-a/b) = m2m1 End of Proof: Q.E.D. |
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