Solution : Distance between (-1/2, -1) , (5/8, -1) = .75
Distance Formula :
Distance between two points can be evaluated if we know the c
[tex]\sqrt{{(y2-y1)^{2} } +(x2-x1)^{2}][/tex] = XY
The difference between the x-axis coordinates gives the horizontal distance and the difference between the y-axis coordinates gives the vertical distance.
Distance between the points P and Q is calculated as follows:
S and T are the points on the x-axis which are endpoints of two parallel line segments PS and QT respectively.
⇒ PR = ST
Coordinates of S and T are (x1, 0) and (x2, 0) respectively.
OS = x1 and OT = x2
ST = OT – OS = x2 – x1 = PR
Similarly,
PS = RT
QR = QT – RT = QT – PS = y2 – y1
By Pythagoras theorem,
PQ2 = PR2 + QR2
PQ = √[(x2– x1)2+ (y2– y1)2]
Therefore,
Distance between two points (x1,y1) and (x2,y2) is given by:
[tex]\sqrt{{(y2-y1)^{2} } +(x2-x1)^{2}][/tex]
distance between (-1/2, -1) , (5/8, -1)
[tex]XY = \sqrt{{(-1-(-1))^{2} } +(5/8-(-1/2))^{2}]\\\\\\\\\\\\XY = \sqrt{9/16}\\\\\\XY = 3/4\\XY = .75\\[/tex]
More details : brainly.com/question/25841655
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