Question

The image of the point (−1, 3, 4) in the plane x − 2y = 0 is

$open parentheses 9 over 5 comma negative 13 over 5 comma space 4 close parentheses$
$open parentheses fraction numerator negative 17 over denominator 3 end fraction comma fraction numerator negative 19 over denominator 3 end fraction comma 1 close parentheses$

(15, 11, 4)

(8, 4, 4)

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Solution

open parentheses 9 over 5 comma negative 13 over 5 comma space 4 close parentheses

If space left parenthesis straight alpha comma space straight beta comma space straight gamma right parenthesis space be space the space image space then space fraction numerator straight alpha minus 1 over denominator 2 end fraction minus 2 space open parentheses fraction numerator straight beta plus 3 over denominator 2 end fraction close parentheses space equals 0 therefore space straight alpha space minus space 1 minus 2 straight beta space minus 6 space space.... space left parenthesis 1 right parenthesis rightwards double arrow space straight alpha space minus 2 straight beta space equals space 7 space.... space left parenthesis 2 right parenthesis and space fraction numerator straight alpha space plus 1 over denominator 1 end fraction space equals space fraction numerator straight beta minus 3 over denominator negative 2 end fraction space equals space fraction numerator straight gamma minus 4 over denominator 0 end fraction from space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space straight alpha space equals space 9 over 5 comma space straight beta space equals space minus space 13 over 5 comma space straight gamma space equals space 4

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