Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest. (There is no diagram available) URGENT!!!
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We have a straight line AB.
Take a point P not on AB, but some distance away. We draw PC the perpendicular on to AB meeting AB at C. We draw another line from P to D, where D lies on AB. We show that PD > PC, for any point D other than C.
In the triangle PCD, use the Pythagorean law for sides:
PC² + CD² = DP²
Since CD² is positive and adds to PC², PC²+CD² > PC²
Hence DP² > PC²
So DP > PC
SO PC, the perpendicular distance from an external point to a line segment, is the shortest.
We have a straight line AB.
Take a point P not on AB, but some distance away. We draw PC the perpendicular on to AB meeting AB at C. We draw another line from P to D, where D lies on AB. We show that PD > PC, for any point D other than C.
In the triangle PCD, use the Pythagorean law for sides:
PC² + CD² = DP²
Since CD² is positive and adds to PC², PC²+CD² > PC²
Hence DP² > PC²
So DP > PC
SO PC, the perpendicular distance from an external point to a line segment, is the shortest.
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97
Answer:
Hey mate refer to the attachment
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