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- 1. STRAIGHT LINES/ FIRST DEGREE EQUATIONS Prepared by: Prof. Teresita P. Liwanag – Zapanta B.S.C.E., M.S.C.M., M.Ed. (Math-units), PhD-TM (on-going)
- 2. STRAIGHT LINES A straight line is a locus of a point thatmoves in a plane with constant slope. It may alsobe referred to simply as a line which contains atleast two distinct points.LINES PARALLEL TO A COORDINATE AXIS If a straight line is parallel to the y-axis, itsequation is x = k, where k is the directeddistance of the line from the y-axis. Similarly, if aline is parallel to the x-axis, its equation is y = k,where k is the directed distance of the line fromthe x-axis.
- 3. The equation of the line through a givenpoint P1 (x1, y1) whose slope is m. y x
- 4. Generally,Considering points P(x, y) and P1 (x1, y1),Therefore, m (x – x1) = y – y1
- 5. The equation of the line having the slope,m, and y-intercept (0, b) y P (x, y) (0, b) b x
- 6. Generally,Considering points P(x, y) and (0, b), mx = y –bTherefore, y = mx + b
- 7. The equation of the line whose x and yintercepts are (a, 0) and (0, b) respectively. y B (0, b) b-y P (x, y) b y A (a, 0) x x a-x a
- 8. ay = -bx + ab
- 9. IV. TWO POINT FORM If the line passes through the points ( x1 , y1 ) and ( x 2 , y 2 ) , then the slope y − y1of the line is m = 2 . Substituting it in the point-slope formula, we have x 2 − x1 y − y1y − y1 = 2 ( x − x1 ) which is the standard equation of the two-point form. x 2 − x1
- 10. The equation of the line through pointsP1 (x1, y1) and P2 (x2, y2) P2(x2, y2) y P (x, y) P1(x1 , y1 ) x
- 11. Considering points P1 (x1, y1) and P2(x2, y2), 1Considering points P(x, y) and P1 (x1, y1), 2 Equation 1 = Equation 2
- 12. Examples:I.Find the general equation of the line:a. through (2, -7) with slope 2/5b. with slope 3 and y-intercept 2/3c. passing through (4, -5) and (-6, 3)d. with x-intercept of 4 and y-intercept of -6e. with slope 1/3 and passing through (5, -3)f. passing through (-2, -7) and has its interceptsnumerically equal but of opposite signs
- 13. g. Determine the equation of the line passingthrough (2, -3) and parallel to the line passingthrough (4,1) and (-2,2).h. Find the equation of the line passing throughpoint (-2,3) and perpendicular to the line2x – 3y + 6 = 0i. Find the equation of the line, which is theperpendicular bisector of the segment connectingpoints (-1,-2) and (7,4).j. Find the equation of the line whose slope is 4and passing through the point of intersection oflines x + 6y – 4 = 0 and 3x – 4y + 2 = 0
- 14. II. The points A(0, 0), B(6, 0) and C(4, 4) arevertices of triangles. Find:a. the equations of the medians and theirintersection pointb. the equations of the altitude and theirintersection pointc. the equation of the perpendicular bisectors ofthe sides and their intersection points
- 15. NORMAL FORM OF THE STRAIGTH LINE A N y C (x1, y1) P y1 x B x1
- 16. Let: AB – given line ON – line perpendicular to AB C– point of intersection with coordinates (x1,y1)Recall: m = tanθ where: m – slope of line θ – Inclination of linemON = tanω therefore, mAB = -1/ tanω mAB = - cotω mAB = - cosω/sinω x1 = Pcosω y1 = Psinω
- 17. DISTANCE FROM A POINT TO LINE y P1 (x1,y1) d P ω x L L1
- 18. Sign Conventions:a. The denominator is given by the sign of B.b. The distance (d) is positive (+) if the pointP1 (x1 ,y1) is above the line.c. The distance (d) is negative (-) if the pointP1 (x1 ,y1) is below the line.
- 19. Examples:1. Find the distance from the line 5x = 2y + 6 to thepointsa. (3, -5)b. (-4, 1)c. (9, 10)2. Find the equation of the bisector of the pair of acuteangles formed by the lines 4x + 2y = 9 and 2x – y = 8.3. Find the equation of the bisector of the acute anglesand also the bisector of the obtuse angles formed bythe lines x + 2y – 3 = 0 and 2x + y – 4 = 0.

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