Machinery's Handbook, 31st Edition
44 LINES AND LINE SEGMENTS If the desired point is the midpoint of AB, then m = n = 1, and the coordinates of P are: x x 1 x 2 + 2 = -------- and y y 1 y 2 + 2 = -------- Example 2: Find the coordinates of a point P that divides the line segment defined by = = (5)(0) + (3)(6) External Division of a Line Segment: If the line segment AB is extended to point Q (Fig. 4c), and Q is such a point that AQ:QB as m:n , then the coordinates of Q are given by: x = and y = mx 1 – nx 2 my 1 – ny 2 m – n m – n A (0, 0) and B (8, 6) at a ratio of 5:3. Solution: x 5 0 × 3 8 × + 5+3 = --------------- 24 8 --- 3 = = y 5+3 --------------- 18 8 --- 2.25 =
Fig. 4. Divisions of Line Segment AB : (a) Midpoint M ; (b) Internal Division Point P ; (c) External Division Point Q . Equation Forms of a Line.— Given any two known points, a linear equation can be ex- pressed in either point-slope or slope-intercept form. Point-Slope Form: Consider that m y 2 y 1 – x 2 x 1 – = -------- for any two points ( x, y ) on a line. If the known point ( x 2 , y 2 ) is replaced with an arbitrary (not specified) point ( x, y ), the slope becomes m y y 1 – x x 1 – = -------- , which rearranges to y – y 1 = m ( x – x 1 ) . This is the point-slope form of the line. If x 1 = y 1 = 0 (that is, the line passes through the origin (0, 0)), then the equation becomes y – 0 = m ( x – 0), or y = mx . For any line, the y -intercept is the point of intersection of the line with the y -axis, and the x -intercept is the point of intersection of the line with the x -axis. Thus, the points (0, y 1 ) and ( x 1 , 0) are the y - and x -intercepts, respectively. Suppose AB intersects the x -axis at point A ( a , 0) and the y -axis at point B (0, b ); then m b a b a = = − − − 0 0 . Substitution of either ( a , 0) or (0, b ) into the formula and rearranging terms gives the equation for AB : y x ay bx ab ay bx ab a − = − + − + − = 0 1 ( ) . Generally, two points are known, and from these, m is determined. Then, either of the points are substituted into the point-slope form along with the slope to get the equation of the line. Example 3: Find the equation of the line that passes through (4, 1) and (–2, 5). So, m = 1 5 4 2 4 6 2 3 − − − − − = = ( ) . Substituting this value for m and the point (4, 1) for ( x 1 , y 1 ) into the point-slope form gives y x − = − − 1 4 2 3 ( ). b a y b x a
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