CHAPTER 6 : COORDINATE GEOMETRY CONTENTS Page 6. Conceptual Map 6. Distance Between Two Points Eercises Division Of A Line Segment 4 Eercises

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ADDITIONAL MATHEMATICS MODULE 0 COORDINATE GEOMETRY

CHAPTER 6 : COORDINATE GEOMETRY CONTENTS Page 6. Conceptual Map 6. Distance Between Two Points Eercises 6. 3 6.3 Division Of A Line Segment 4 Eercises 6.3. 5 Eercises 6.3. 7 SPM Questions 8 Assessment 0 Answer

6. Conceptual map COORDINATE GEOMETRY Distance between two points. Division of a line segment 6. Distance between two points. Note: y y B(, y ) y y Distance AB = y A(, y ) Eamples : Solution.. Find the distance between A(,3) and B(7,6). Use (, y ) (5, 3) and (, y ) = (8, 7). Therefore, AB = ( 8 5) (7 = = = 5 units 3)

Eamples : Solution.. Given that the distance between R(4, m) and S(-, 3) is 3 units, find the value of m. Given that RS = 3 units, therefore ( 4) (m 3) = 3 = (m 3) = m 3 = = or = m = or m = EXERCISES 6.:. Find the distance between each of the following pairs. (a) C(, 3) and D(4,-) (b) R(-, 6) and T(7, -3) (c) K(-5, -) and L(-6,-) (d) P(, ) and Q(, -3). Given the points R(-9, -), S(-, -6) and T(-,),show that TR = RS. 3

3. The distance between the points U(6,3t) and the points V(,-t) is 0 units. Find the possible value of t. 4. Given point P(h, k) is equidistant from points A(, 5) and B(-, 4). Show that k + 8h = 9. 6.3 Division of A Line Segment 6.3. The midpoint of two points. Note: y y B(, y ) y A(, y ) C(,y) If C is the midpoint of the line AB, then coordinate C = Eamples :. Find the midpoints of points P(3, 4) and Q(5, 8) Solution. 3 5 4 8 (a) Midpoint PQ =, 8 =, = (4,6). 4

Eamples :. Points A and B are (5, r) and (, 7) respectively. Find the value of r,if the midpoint of AB = (3, 5). Solution. 5 r 7 Midpoint of AB =, (3, 5) = 3, r 7 r = 3. B(3, 4), C(7, 5), D(6, ) and E are the vertices of a parallelogram. Find the coordinates of the point E Let the verte E be (, y). 3 6 The midpoint of BD =, =, 3. 5 y The midpoint of CE =, The midpoint of BD = The midpoint of CE. 9, =,3 7 9 5 y and 3 = and y =. EXERCISES 6.3.. Find the midpoint of each pair of points. (a) I(4, -5) and J(6, 3) (b) V(-4, 6) and W(,-0) 5

. If M(3, q) is the midpoint of the straight line K(, 6) and L(4, 5). Find the value of q. 3. The coordinates of points R and S are (4, k) and (h,5) respectively. Point T(-, ) is the midpoint of RS. Find the values of h and k. 6.3. Finding the coordinates of a point that divides a line according to a given ratio m : n. Note: y n Q(, y ) If A divides PQ according to a ratio m : n,then m A A(, y) = P(, y ) Eamples. Given that G(-3, 6) and H(7, ). If B divides GH according to the ratio :3, find the coordinates of B. 6 Solution Let the coordinates of point B be (,y). n m ny my Coordinates of point B =, m n m n 3( 3) (7) 3(6) () =, 3 3 = (, 4).

Eamples. Given the points P(-, 3) and Q(8, 9). Point R lies on the straight line PQ such that PR = RQ. Find the coordinates of point R. Solution PR Given PR = RQ, therefore, RQ m = and n = ( ) (8), (3) (9) Coordinates of point R = = (, 5) 3. Point P(-3, -) divides internally a line segment joining two points R(6, ) and S(-6, -3). Find the ratio of division of line segment RS. Let the ratio is (m, n). n(6) m(-6) n() m(-3) (-3,-) =, m n m n 6n 6m -3 = m n -3(m + n) = 6n + 6m -3m 3n = 6n + 6m 3m = 9n m 3, m : n = 3 :. n EXERCISES 6.3.. The coordinates of K and L are (0, 5) and (5, 5) respectively. If point M divides KL to the ratio of : 3, find the coordinates of point M.. Given point P(k, ), Q(0,3) and R(5, 4) find the possible values of k if the length of PQ is twice the length of QR. 7

3. P(a, ) is a point dividing the line segment joining two points A(4, 3) and B(-5, 0) internally in the ratio m : n. Find (a) m : n. (b) the value of a. 4. K(-4, 0), L and P(8, 6) are three points on the straight line KL such that coordinates of point L in terms of m. KL m. Find the LP SPM QUESTIONS.. The points A(h, h), B(p, t) and C(p, 3t) are on a straight line. B divides AC internally in the ratio : 3. Epress p in the terms of t. (003, Paper ) 8

. Diagram shows a straight line CD meets a straight line AB at the point D. The point C lies on the y- ais. (004/P) y C 0 B(9, 0) A(0, -6) D Given that AD = DB, find the coordinates of D. 9

ASSESSMENT (30 minutes). Given the distance between point Q (4, 5) and R (, t) is 5, find the possible values of t.. Given the points A (-, 3), B (-4, 7) and C (5, -6).If P is the midpoint of AB, find the length of PC. 3. In the diagram, PQR is a straight line. Find the ratio of PQ : QR. R(, 7) Q(-, ) P( 5, ) 4. The points P(h, h), Q( k, l ) and R( 3k, l ) are collinear. Q divides PR internally in the ratio 3 :. Epress k in the terms of l. 0

ANSWERS: Eercise 6.. (a) 5 units (b).78 units (c).44 units (d) 4.03 units 3., - units. Eercise 6.3.. (a) ( 5, 4 ) (b) (-, -). 5 3. h = -6, k = -. Eercise 6.3.. (8, 9). -0 3. (a) : (b) - 8m 4 6m 4., m m SPM QUESTIONS. p = -t. D = (3, -4) ASSESSMENT., 9. 3.605 units. 3. : 4. m n 8

ADDITIONAL MATHEMATICS MODULE COORDINATE GEOMETRY

CONTENTS page 6. Concept map... 6. Gradient of a straight line...3 6... Ais interceptions...3 Eercise...3-4 6... Gradient of a straight line that passes through two points...4 Eercise...5 6.3 Self assessment...5-7 6.4 Equation of a straight line...7 6.4.. Given the gradient and passing through point (,y)...8 6.4.. Line which passed through two points...8 6.4.3 Given the gradient and y-intercept...9 6.5 Self assessment...9-0 6.4.4. Point of intersection of two lines... 6..6 Self assessment 3... Short test... Answers...3

CHAPTER 6: COORDINATE GEOMETRY 6. : CONCEPT MAP GRADIENT TWO POINTS AXIS INTERCEPTIONS EQUATIONS OF A STRAIGHT LINE GRADIENT FORM INTERCEPT FORM GENERAL FORM INTERSECTIONS OF TWO LINES SIMULTANEOUS EQUATIONS ELIMINATION METHOD SUBSTITUTION METHOD 3

6..GRADIENT OF A STRAIGHT LINE 6.. AXIS INTERCEPTIONS Find the -intercepts,y-intercepts and gradients of the following straight lines PQ. Eample: y P(0,5) (, y ) y-intercept=5 Q(3,0) (, y ) -intercept=3 GradientPQ y y 0 5 = 3 0 = 3 5 EXERCISE : 4

.Find the gradients of the following straight lines.. -3 y y Gradient=.... y -3 5 Gradient =... 3.Find the gradient of a straight line that intercepts the -ais at and the y-ais at -4 Answer: 4. y A - The diagram shows a straight line which has a gradient of ½. Find the coordinates of point A. Answer: 6.3.Gradient of a straight line that passes through two points 5

y Gradient = y Eample: Find the gradient of a straight line that passes through points A (,-4) and B(4,8) Solution: 8 ( 4) Gradient of AB, m AB = 4 = 6 EXERCISE. Find the gradient of the straight line joining each of the following pairs of points (a) (,3) and (4,9) Answer: (b) (-,) and (,8) Answer:. Find the value of h if the straight line joining the points (h,-3) and (-,-h + ) has a gradient of. Answer: 3.Given P(3a-,-a), Q(-5,3) and R(,6) are three points on a straight line. Find the value of a. Answer: 6

.Find the gradients of the following straight lines. (a) y y (b) 0 y 3 Gradient =... (c) Gradient=... 4 (d) 4 Gradient =... Gradient =....The diagram shows a straight line which has a gradient of. Find the coordinates of point B. Answer... y B - 7

3.A straight line with a gradient of ½ passes through a point (-,4) and intersects the -ais and the y- ais at points A and B respectively. Find the coordinates of points A and B. Answer: 4.A straight line has a gradient of h/4 and passes through a point (0,4h). (a) Find the equation of the straight line. (b) If the straight line passes through point (-4,3),find the value of h. Answer: (a) (b) 5.. Y C B(-,4) X A(,) The figure on the previous page shows a triangle ABC with A(,) and B(-,4). The gradients of AB, AC and BC are -3m,3m and m respectively. (i) Find the value of m (ii) Find the coordinates of C (iii) Show that AC=AB Answer: (i) (ii) 8

(iii) ************************************************************************ 6.4 EQUATION OF A STRAIGHT LINE METHODOLOGY: 6.4.. GIVEN THE GRADIENT (m) AND PASSING THROUGH POINT ( X, Y ) 6.4. LINE WHICH PASSED THROUGH TWO POINTS ( X, Y ) AND ( X, Y ) 6.4.3 GIVEN THE GRADIENT(m) AND Y-INTERCEPTS(c) 6.4.GIVEN THE GRADIENT(m) AND PASSING THROUGH POINT X, Y ) ( m (,y) Y-y=m(-) Y=m + c Equation in general form a+by+c=0 Eamples: 5 (0,-4) Y-(-4)=5(-0) Y+4 =5 Y=5-4 5-y-4=0.. 3 3 (3,0) (-3,0) 3. -3 (-,4) 6.4. LINE WHICH PASSED THROUGH TWO POINTS X, ) AND X, Y ) ( Y ( y y Eample 3 4 9 y y y y Y=m+c Equation in general form (a+by+c=0) y 3 9 3 4 9

y 3 6 3 y-3=(-) y-3=- Y=-+3 Y=+ -y+=0. - -8. - -b 4b 3. 3a- -a -5 3 6.4.3 GIVEN THE GRADIENT(m) AND Y-INTERCEPTS(c) m y-intercept Y=M+c a b y A+by+c=0 Eample: 3-3 Y= 3 y 3 3-3-y-6=0 5-4 - -6 3-5 - 0

************************************************************************.A straight line which passes through the points (,3) and (5,m) has gradient m. Find the value of m. Answer:.Find the equation of the straight line which joins the points P(-3,4) and Q (-,-). the line intersects the y-ais at the point R, find the length of PR Answer: Given that 3.Given the points A (-,5), B(,7) and C (4,0). The point P divides the straight line BC in the ratio :. Find the equation of the straight line which passes through the points P and A. Answer: 4.The straight line intersects -ais and y-ais at point A(3,0) and B(0,0 respectively. Find (a) the equation of straight line in: (i) Intercept form Answer (ii) Gradient form Answer

(iii) general form Answer: (b) the equation of straight line that passes through the point A and with gradient Answer: 6.4.4 Point of intersection of two lines eample: Find the point of intersection of the straight lines y=- + and y = 6 Solution: y = - +...() y = 6...() () 4: 4y = + 4...(3) () + (3): 5y = 5 y = 5 Substitute y =5 into () 5 = - + = -4 = - Therefore the point of intersection is (-,5) ************************************************************************.Two straight lines y and ky = - + intersect the y-ais at the same point. Find the value 6 of k Answer;

.A straight line passes through a point (5,) and the -intercept is 0. If the straight line intersects the y-ais at point R, find (a) the equation of the straight line Answer (b) the coordinate of point R Answer ****************************************************************************** ******************************************************************************. The diagram below shows a straight line CD which meets a straight line AB at point C. Point D lies on the y-ais. (a) Write down the equation of AB in the intercept form (b) Given that AC = CB, find the coordinate of point C Answer: (a) (b) y C B(0,3) 3

A(-6,0) 0 D Eercise. ) /3 ) 3/5 3) 4) 4 Eercise a) m= b) m=3 ) h= -3 3.) a= - Self assessment I a) 3 b) 3 c.) zero d) undefined ) A(4,0) 3) A(-0,0), B(0,5) 4a) h y 4h 4 4b) h= 5(i) m=½ (ii) C=(5,7) (iii) AC= 5, AB 3 Self assessment II ) m=-3/ )PR= 90unit 3) y+=44 4

4a.(i) y 3 (ii) y 3 (iv) --3y + 6 =0 4b) y=-6 Self assessment III ) k= a) 5y=-+0 b) R(0,) Short test y a) 6 3 b) C(-4,) 5

ADDITIONAL MATHEMATICS MODULE COORDINATE GEOMETRY CONTENTS Subtopic Page 6.0 Parallel lines and perpendicular lines 6. Conceptual Map 6.. Parallel Lines Eercise 6.. 3 6.. Perpendicular Lines 4 Eercise 6.. 5 6..3 Problem Involving The Equation of Straight Line 6 Eercise 6.3.. 6 6

6.3 Equation of a Locus 8 6.3. Locus of a point that moves in such away that its distance 8 from a fied points is a constant Eercise 6.3. 8 6.3. Locus of a point that moves in such away that the ratio of 9 its distance from fied points is a constant. Eercise 6.3. 9 6.3.3 Problem Solving Involving Loci 0 Eercise 6.3.3 6.4 Area of Polygon 6.4. Area of Triangle 6.4. Area of Quadrilateral Eercise 6.4 4 SPM Question 5 Assessment 6 Answer 7 6.0 PARALLEL LINES AND PERPENDICULAR LINES. 6. Conceptual Map COORDINATE GEOMETRY Lines Equation of locus involving distance between two points. Lines Triangle Quadrilateral Area of polygon 7

6.. Parallel line. Notes: Two straight line are parallel if m m and vise versa. Eamples. Determine whether the straight lines y = 5 and y = 3 are parallel. Solution y = 5, y = 5, m y = 3 y = 3, m Since m m, therefore the straight lines y = 5 and y = 3 are parallel.. Given that the straight lines 4 + py = 5 and 5y 6 = 0 are parallel, find the value of p. Step: Determine the gradients of both straight lines. 4 + py = 5 4 5 4 y =, m p p p 5y 6 = 0 5 y = 3, m 5 Step : Compare the gradient of both straight lines. Given both straight lines are parallel, hence m m 4 p 5 p = -0 3. Find the equation of the straight line which passes through the point P(-3, 6) and is parallel to the straight line 4 y + = 0. 4 y + = 0, y = +. Thus, the gradient of the line, m =. Therefore, the equation of the line passing through P(-3, 6) and parallel to the line 4 y + = 0 is y - 6 = ( - -3) y = +. EXERCISES 6... 8

. Find the value of k if the straight line y = k + is parallel to the straight line 8 y + = 0.. Given a straight line 3y = m + is parallel to y. 3 5 Find the value of m. 3. Given the points A(, ), B(4, -3),C(5, 4) and D(h, -). If the straight line AB is parallel to the straight line CD, find the value of h. 4. Find the equation of a straight line that passes through B(3, -) and parallel to 5 3y = 8. 5. Find the equation of the straight line which passes through the point A(-, 3) and is parallel to the straight line which passes through the points P(, ) and Q(5, ). 9

6.. Perpendicular Lines. Notes: Two straight lines are perpendicular to each other if m m and vise versa. Eamples. Determine whether the straight lines 3y = 0 and y + 3 + 4 = 0 are perpendicular. Solution 3y = 0 y =, 3 3 m 3 y + 3 + 4 = 0 y= 3 4, m 3 m m ( 3) = -. 3 Hence, both straight lines are perpendicular. Eamples. Find the equation of the straight line which is perpendicular to the straight line + y 6 = 0 and passes through the point (3, -4). Solution + y 6 = 0 y = 3, m Let the gradient of the straight line which is perpendicular = m m = - m = The equation of the straight line = y = 0

EXERCISES 6.... The equation of two straight line are y 3 5 perpendicular to each other. and 3 5y = 8. Determine whether the lines are. Find the equation of the straight line which passes through point (, 3) and perpendicular of the straight line y + = 4. 3. Given the points A(k, 3), B(5, ), C(, -4) and D(0, 6). If the straight line AB is perpendicular to the straight line CD, find the value of k. 4. Find the value of h if the straight line y h + = 0 is perpendicular to the straight line 5y + + 3 = 0

6..3 Problem involving The Equations Of Straight Lines. Eamples. Given A(3, ) and B(-5, 8). Find the equation of the perpendicular bisectors of the straight line AB. The gradient of AB, Solution m The gradient of the perpendicular line = m m m m = The midpoint of AB = The equation of the perpendicular bisector, = EXERCISES 6.5.4. Given that PQRS is a rhombus with P(-, ) and R(5, 7), find the equation of QS.. ABCD is a rectangle with A(-4, ) and B(-, 4). If the equation of AC is 4 + 7y + = 0, find (a) the equation of BC. (b) the coordinates of points C and D. 3. PQRS is a rhombus with P(0, 5) and the equation of QS is y = +. Find the equation of the diagonal PR.

4. A(, k), B(6, 4) and C(-, 0) are the vertices of a triangle which is right-angled at A. Find the value of k. 6.3 EQUATION OF A LOCUS 6.3. Locus of a point that moves in such a way that its distance from a fied point is a constant. Eample:. A point K moves such that its distance from a fied point A(,) is 3 units. Find the equation of the locus of K. Solution Let the coordinates of K be (,y) Distance of KA = Hence, = 3 unit ( ) ( y ) = 4 4 y 4 4 y = y 9 y 9 0 y 4 y 4 0 The equation of the locus of P is y 4 y 4 0 3

Eercises 6.3.. Find the equation of the locus point M which moves such that its distance from each fied point is as follows. a. 5 units from A (-,) b. 7 units from B (-3,-) c. units from C (0,) d. 3 units from D (,0) 6.3. Locus of a point that moves in such away that the ratio of its distances from two fied points is a constant. Eample: A point P moves such that it is equidistant from points A (,-) and B (3,). Find the equation of the locus of P. Solution: Let the coordinates of P be (,y) Distance of AP = Distance of BP y 3 y 4 4 y y 6 9 y + 6y = 8 + 3y = 4 The equation of the locus of P is + 3y =4 4y 4 4

Eercises 6.3.. A point P moves such that it is equidistant from points A (3,) and B (,). Find the equation of the locus of P.. Given points R(4,), S(-,0) and P (,y) lie on the circumference of diameter RS. Find the equation of the moving point P. 3. Points A(4,5),B(-6,5) and P are vertices of a triangle APE. Find the equation of the locus of point P which moves such that triangle APB is always right angled at P. 5

6.3.3 Problems solving involving loci Eample:. Given points A (, 5), B (-6,-) and P(, y) lie on the circumference of a circle of diameter AB. Find the equation of the moving point P. Solution: Distance of AB = ( 6) (5 0) = = 0 units So radius of circle = Let the mid-point of points A and B be C. C is also the center of the circle. ( 6) 5 ( ) Coordinates of C =, = (-,) Hence, CP = 5 y = 5 4 4 y 4y 4 5 y 4 4y 7 0 The equation of the locus of P is Eercise 6.3.3. A point P moves along the arc of a circle with center C (3,). The arc passes through A(0,3) and B (7,s). Find (a) the equation of the locus of point P (b) the values of s 6

. Given the points are A (,-) and B(,-). P is a point that moves in such a way that the ratio AP: BP = : (a) Find the equation of the locus of point P (b) Show that point Q (0,-3) lies on the locus of point P. (c) Find the equation of the straight line AQ (d) Given that the straight line AQ intersects again the locus of point P at point D, find the coordinate of point D. 6.4 AREA OF POLYGON. 6.4. Finding the area of triangle Note: y A(, y ) C, y 3 ) ( 3 B, y ) ( Area of triangle ABC = = 7

Area of triangle ABC = 0 A, B and C are. If the coordinate of the vertices are arranged clockwise in the matri form, the area of triangle obtained will be a value. 6.4. Finding the area of quadrilateral Given a quadrilateral with vertices A( ), B( ), C( ) and D(, ). The area of quadrilateral ABCD = =, y, y, y 3 3 ( 4 y4 Eample. Given P(-, 3), Q(3, -7) and R(5, ), find the area of triangle PQR. y P(-,3) Solution R(5, ) Q(3, -7) Area of triangle PQR 3 5 = 3 7 3 7 3 5 9 ( 35) ( ) = (5) = 6 units. =. The vertices of a triangle are (-3, q), (5,) and (-, 3). If the triangles has an area of 0 unit, find the possible values of q. Area of triangle = 5 3 3 q 5 = 0 8

5 ( q) ( 3) ( ) ( 9) 5q = 0 6q = 40 6q = 40 or 6q = 40 q = -3 or q = 0. 3 3. ABCD is a parallelogram. Given A (-, 7), B(4, -3)and C(8, -), find (a) point D (b) the area of the parallelogram. (a) Let the verte D be (, y). 8 7 ( ) The midpoint of AC =, = (3, -) The midpoint of BD = 4 3, y The midpoint of BD = The midpoint of AC. 4 3, y = (3, -) 4 3 y 3 and = and y = -. Point D(, -). (b) The area = = (6) = 8 units. 7 8 4 3 = ( ) ( 4) (8) 4 ( 8) ( 44) 6 7 EXERCISES 6.4. Given S(, ), T(0,7) and U(5,4) are the vertices of STU. Find the area of STU. 9

. Find the possible values of k if the area of a triangle with vertices A(3, ), B(-, 6) and C(k, 5) is 8 units. 3. Show that the points (-9, ), (3, 5) and (, 7) are collinear 4. The vertices of quadrilateral PQRS are P(5, ), Q(a, a), R(4, 7) and S(7, 3). Given the area of quadrilateral PQRS is unit, find the possible values of a. 5. Given that the area of the quadrilateral with vertices A(5, -3), B(4, ), C(-3, 4) and D(p, q) is 9 unit, show that 7p + 8q 6 = 0. SPM QUESTIONS. 30

. The equations of two straight lines are y 5 3 lines are perpendicular to each other. and 5y = 3 + 4. Determine whether the (003/P). Diagram shows a straight line PQ with the equation y 3. The points P lies on the - ais and the point Q lies on the y ais. (004/P) y Q 0 P Find the equation of a straight line perpendicular to PQ and passing through the point Q. 3. A point P moves along the arc of a circle with centre A(, 3). The arc passes through Q(-, 0) and R(5,k). (003/P) (a) Find the equation of the locus of the point P, (b) Find the values of k. 3

5. The point A is (-, 3) and the point B is (4, 6). The point P moves such that PA:PB = :3. Find the equation of the locus P. ASSESSMENT (30 minutes) 5. Find the equation of the straight line which is parallel to line y + = 7 and passes through the point of intersection between the lines 3y = and y = 3. 6. Given A(6, 0) and B(0,-8). The perpendicular bisector of AB cuts the aes at P and Q. Find (a) the equation of PQ, (b) the area of POQ, where O is the origin. 7. The point moves such that its distance from Q(0, 4) and R(, 0) are always equal. The point S however moves such that its distance from T(, 3) is always 4 units. The locus of point P and the locus of point S intersect at two points. (a) Find the equation of the locus of the point P. (b) Show that the locus of the point S is + y 4 6y 3 = 0. (c) Find the coordinates of the points of intersection for the two loci. 3

8. Find the possible values of k if the area of a triangles with vertices A(9, ), B(4, ) and C(k, 6) is 30 units. Answer: Eercise 6... k = 4. m = -5 3. h = 8 4. 5 y 6 5. 3 y 4 5 Eercise 6... perpendicular to each other 33

. y = + 3. k = 5 4. h = 5 Eercise 6..3. + y 6 = 0. (a) 3 + y 5 = 0 (b) C(3, -), D(0, -4) 3. + y 0 = 0 4. k = or k =. Eercise 6.3.. a. y 4 y 0 b. y 6 y 3 0 c. y y 43 0 d. y 4 5 0 Eercise 6.3.. + y = 4. y y 0 3. y 0y 0 8 Eercise 6.3.3 a. y 6 y 5 0 b. s = 4 @ s = - a. 3 3y 4 4y 5 0. b. substitute = 0, y = 3 c. y = 3 4 5 d. D (, ) 3 3 Eercise 6.4 9. units. k = -4, 4 6 3. a = 8 or a= 7

SPM QUESTION. Perpendicular. y 3 3 3. + y 4 6y = 0 4. k = - or k = 7. ASSESSMENT. y + + 7 = 0. (a) 3 + 4y + 7 = 0 (b) unit 4 3. (a) y + 3 = 0 (c) = 5.76, -.36 y = 4.38, 0.8