Find the value of x for which the distance between the point p(9,-5)and q(12,x)is 3 unit
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(Dis)² = ( x2 - x1) ² + ( y2 - y1) ²
9 = ( 12 - 9 )² + (x +5 ) ²
9 = 9 + x² + 25 + 10x
x² + 10x + 25 = 0
x² + 5x + 5x + 25 = 0
x(x + 5) + 5(x + 5) = 0
(x +5 ) + (x + 5) = 0
Therefore value of x is -5
9 = ( 12 - 9 )² + (x +5 ) ²
9 = 9 + x² + 25 + 10x
x² + 10x + 25 = 0
x² + 5x + 5x + 25 = 0
x(x + 5) + 5(x + 5) = 0
(x +5 ) + (x + 5) = 0
Therefore value of x is -5
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