If RS=44 and QS=68, Find QR
Line QS is a continuation of line RS : Q R = Q S + Q R = 68 + 44 = 112 QR = QS + QR = 68 + 44 = 112 QR = QS + QR = 68 + 44 = 112 Point R belongs to line QS: Q R = Q S − R S = 68 − 44 = 24 QR = QS - RS = 68 - 44 = 24 QR = QS − RS = 68 − 44 = 24
Find the distance between the point (1,4) and (-2,-1):
Distance = ( − 2 − 1 ) 2 + ( − 1 − 4 ) 2 = ( − 3 ) 2 + ( − 5 ) 2 = ( 9 + 25 ) = 34 \sqrt{( -2 - 1)^2 + (-1 - 4)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{(9 + 25)} = \sqrt{34} ( − 2 − 1 ) 2 + ( − 1 − 4 ) 2 = ( − 3 ) 2 + ( − 5 ) 2 = ( 9 + 25 ) = 34
Find the midpoint of segment with endpoints (9,8) and (3,5):
Midpoint(xc , yc ):
x c = x a + x b 2 = 9 + 3 2 = 12 2 = 6 , y c = y a + y b 2 = 5 + 8 2 = 13 2 = 6.5 x_c = \dfrac{x_a + x_b}{2} = \dfrac{9 + 3}{2} = \dfrac{12}{2} = 6 ,
y_c = \dfrac{y_a + y_b}{2} = \dfrac{5 + 8}{2} = \dfrac{13}{2} = 6.5 x c = 2 x a + x b = 2 9 + 3 = 2 12 = 6 , y c = 2 y a + y b = 2 5 + 8 = 2 13 = 6.5
Midpoint(6, 6.5)
ABC is right Triangle AB =
Pythagorean theorem A B = B C 2 + A C 2 AB = \sqrt{BC^2 + AC^2} A B = B C 2 + A C 2 ( i f ∠ С = 90 ° ) ( if \angleС = 90\degree) ( i f ∠ С = 90° )
Find the distance between the points (1,-8)and(-7,-2):
Distance = ( − 7 − 1 ) 2 + ( − 2 − ( − 8 ) ) 2 = ( − 8 ) 2 + ( − 2 + 8 ) 2 = ( − 8 ) 2 + ( 6 ) 2 = ( 64 + 36 ) = 100 = 10 \sqrt{( -7 - 1)^2 + (-2 - (-8))^2} = \sqrt{(-8)^2 + (-2 + 8)^2} = \sqrt{(-8)^2 + (6)^2} = \sqrt{(64 + 36)} = \sqrt{100} = 10 ( − 7 − 1 ) 2 + ( − 2 − ( − 8 ) ) 2 = ( − 8 ) 2 + ( − 2 + 8 ) 2 = ( − 8 ) 2 + ( 6 ) 2 = ( 64 + 36 ) = 100 = 10
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If RS =32.5 and QS= 62.1 find QR