Prove: The product of the slopes of lines AC and BC is -1.

Mathematics

1 answer:

kicyunya [14]2 years ago

6 0

Given that the AC and BC are perpendicular, their slopes are the negative inverse of each other which gives that the product of the slopes of AC and BC is -1

<h3>How can the prove that the product of the slopes is -1 be found?</h3>

The completed proof is presented as follows;

The slope of AC or GC is GF/FC by definition of slope. The slope of BC or CE is DE/CD by definition of slope.

<FCD = <FCG + <GCE + <ECD <u>by angle addition property </u> <FCD = 180° by the definition of a straight angle, and <GCE = 90° by definition of perpendicular lines. So by substitution property of equality 180° = <FCG + 90° + <ECD. Therefore 90° - <FCG = <ECD, by the <u>subtraction property of equality </u> . We also know that 180° = <FCG + 90° + <CGF by the triangle sum theorem and by the subtraction property of equality 90° - <FCG = <CGF, therefore <ECD = <CGF by the substitution property of equality. Then <ECD ≈ <CGF by the definition of congruent angles. <GFC ≈ <CDE because all right angles are congruent. So by AA ∆GFC ~ ∆CDE. Since <u>the ratio of corresponding sides of similar triangles are equal</u> then GF/CD = FC/DE or GF•DE = CD•FC by cross product. Finally, by the division property of equality GF/FC = CD/DE. We can multiply both sides using the slope of using the <u>multiplication property of equality </u> to get GF/FC × -DE/CD = CD/DE × -DE/CD. Simplify so that GF/FC × -DE/CD = -1. This shows that the product of the slopes of AC and BC is -1.