We used the similar triangles to show that the product of...

We used the similar triangles to show that the product of the slopes of two perpendicular lines equals -1 . The steps below outline an alternative proof that avoids the use of $\operatorname{sim}-$ ilar triangles but uses more algebra instead. Use the figure below, which is the same as the figure used earlier except that there is now no need to label the angles. (a) Apply the Pythagorean Theorem to triangle $P S Q$ to find the length of the line segment $P Q$ in terms of $a$ and $b$. (b) Apply the Pythagorean Theorem to triangle $P S T$ to find the length of the line segment $P T$ in terms of $a$ and $c$. (c) Apply the Pythagorean Theorem to triangle $Q P T$ to find the length of the line segment $Q T$ in terms of the lengths of the line segments of $P Q$ and $P T$ calculated in the first two parts of this problem. (d) As can be seen from the figure, the length of the line segment $Q T$ equals $b+c$. Thus set the formula for length of the line segment $Q T$, as calculated in the previous part of this problem, equal to $b+c,$ and solve the resulting equation for $c$ in terms of $a$ and $b$. (e) Use the result in the previous part of this problem to show that the slope of the line containing $P$ and $Q$ times the slope of the line containing $P$ and $T$ equals -1 .

We used the similar triangles to show that the product of the slopes of two perpendicular lines equals -1 . The steps below outline an alternative proof that avoids the use of $\operatorname{sim}-$ ilar triangles but uses more algebra instead. Use the figure below, which is the same as the figure used earlier except that there is now no need to label the angles.
(a) Apply the Pythagorean Theorem to triangle $P S Q$ to find the length of the line segment $P Q$ in terms of $a$ and $b$.
(b) Apply the Pythagorean Theorem to triangle $P S T$ to find the length of the line segment $P T$ in terms of $a$ and $c$.
(c) Apply the Pythagorean Theorem to triangle $Q P T$ to find the length of the line segment $Q T$ in terms of the lengths of the line segments of $P Q$ and $P T$ calculated in the first two parts of this problem.
(d) As can be seen from the figure, the length of the line segment $Q T$ equals $b+c$. Thus set the formula for length of the line segment $Q T$, as calculated in the previous part of this problem, equal to $b+c,$ and solve the resulting equation for $c$ in terms of $a$ and $b$.
(e) Use the result in the previous part of this problem to show that the slope of the line containing $P$ and $Q$ times the slope of the line containing $P$ and $T$ equals -1 .

Key Concepts

Algebraic Manipulation in Geometric Proofs

Many geometric problems, including those involving perpendicularly and similar figures, often require the translation of geometric relationships into algebraic equations. The process of forming and solving these equations is key in establishing relationships between various geometric quantities.

Distance Formula

Derived from the Pythagorean Theorem, the distance formula calculates the distance between two points in the coordinate plane. It provides a method to measure lengths of segments which is fundamental when applying algebraic techniques to geometric problems.

Pythagorean Theorem

This fundamental theorem in Euclidean geometry relates the lengths of the sides in a right triangle. It states that the square of the hypotenuse is equal to the sum of the squares of the other two sides, a concept frequently used to compute distances in geometry when right angles are involved.

Slope of a Line

The slope is a measure of the steepness or incline of a line in the coordinate plane, defined as the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line. It is a central concept in analytic geometry for describing line behavior.

Perpendicular Lines and Negative Reciprocals

In coordinate geometry, two lines are perpendicular if the product of their slopes is -1, meaning that the slope of one is the negative reciprocal of the other. This property is essential for proving perpendicularity and is widely used in both geometric proofs and practical applications.

Transcript

00:05 So according to the given figure, i have three points basically o, p, that is x1, m1, and another point q, that is x2 comma m2x2.

00:20 So in the first part, if i have to find the distance of dop, so that distance will be written as under root of x1 minus x2 square, so that will be as x1 minus 0 square, that is x1 square only plus m1 x1 minus 0 whole square that is m1 square x1 square similarly d of o q in the next part if i have to get it that will be x2 minus 0 square that is x2 square plus m2 x2 minus 0 whole square that is m2 square x2 square finally d of pq that will be obtained as under root of x2 x1 minus x2 whole square plus m1 x1 minus m2 x2 whole square so that is the expression for pq now if i have to get this triangle a right angle triangle then in the third part uh sorry fourth part i will have to tell that d of pq basically i have d of pq square is equal to d of opi square plus d of oq square so when i substitute this values and try to simplify squaring this will remove the under root expression so i can directly write in left hand side x1 square plus x2 square minus 2 x1 x2 plus m1 square x1 square plus m2 square x2 square minus 2 m 1 m 2 x1 x2 that is equal to d of opi square that is x1 square plus m1 x1 square plus the next one x2 square plus m2 square x2 square so i can see clearly that x1 square x2 square gets cancelled and m1 square x1 square m2 square x2 square all gets cancelled so ultimately what am i left out with is minus 2 x1 x2 minus 2 m1m 2 and x1 x2 all multiplied that is equal to 0 so that is my fourth part now in the fifth part they're asking me to take minus 2 x1 x2 common as a factor so then i'll be left out with 1 plus m1 m2 is equal to 0 now using the 0 property the 0 product.

03:33 So using the 0 product property, i will say that either x1 x2 equal to 0 or 1 plus m1m2 is equal to 0.

03:44 So x1x2 are clearly points away from the origin.

03:49 They are distinct points.

03:50 So this cannot be true...

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