Parallel lines have the same slope.
The given line has a slope of 3, therefore, the line parallel to the given line has a slope of 3.
Now, we have the slope of the line and a point that it passes through.
So, we can use the Point-Slope formula:
Answer:
The 7 in the one's place is 100 times the value of the 7 in the hundreds place.
Do you mean like equal groups
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.
Learn more about perpendicular lines here:
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Subtract 612 from 8.7 to get 603.3 :)
