Equation of a Line. Chapter H2. The Gradient of a Line. m AB = Exercise H2 1

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1 Chapter H2 Equation of a Line The Gradient of a Line The gradient of a line is simpl a measure of how steep the line is. It is defined as follows :- gradient = vertical horizontal horizontal A B vertical The letter we use to define the gradient is m. In fact, if we talk about the gradient of the line joining A and B, we use the notation :- m AB = vertical horizontal At Higher Level, there is a more mathematical wa of epressing this gradient formula :- Given that A( 1, 1 ) and B( 2, 2 ) are an two points in a coordinate diagram, (as shown opposite), => m AB = => m AB = vertical horizontal = BC AC A( 1, 1 ) 1 2 ( 2 1 ) B( 2, 2 ) ( 2 1 ) C 1 2 Eample 1 :- Find the gradient of the line joining A( 1, 3) to B(7, 5). Here, 1 = 1, 2 = 7, 1 = 3, 2 = 5. Eample 2 :- Find the gradient of the line joining P( 5, 8) to B(1, 2). Here, 1 = 5, 2 = 1, 1 = 8, 2 = 2. m = = ( 1) = 2 8 = 1 4 m = 2 1 = ( 2) ( 5) = 10 6 = 5 3 Positive gradient => line slopes upwards Negative gradient => line slopes downwards Eercise H Given that C(3, 1) and D(7, 4), cop and complete the following to find m CD. Here, 1 = 3, 2 =..., 1 = 1, 2 =... => m = 2 1 = CD 2 1 (... 1) (... 3) = Calculate the gradient of the line joining R( 1, 2) to S(1, 8). (i.e. find m RS.) = Given G(1, 9) and H(7, 3), find m GH. What does our answer tell ou about the line GH? 4. Use the gradient formula shown in Eamples 1 and 2 above to determine the gradients of the lines joining the following pairs of points :- (a) A(2, 1), B(8, 3) (b) C(0, 3), D(10, 8) (c) E(1, 1), F(3, 5) (d) G( 1, 3), H(2, 4) (e) I(4, 1), J( 1, 3) (f) (K(2, 3), L( 2, 5). 5. On a small Cartesian diagram, plot the two points R(2, 1) and S(6, 1). (a) Calculate the gradient m RS. (b) If the gradient of an line is zero, what can ou sa about the line? Chapter H2 this is page 209 Equation of a Line

2 6. This time, draw a new diagram and plot the two points U(4, 1) and V(4, 3). Calculate the gradient of the line UV. (note :- it is NOT zero! - Use our calculator if ou are not sure of the value). (We sa that the gradient of a vertical line is infinite or more commonl, it is said to be undefined - too steep to be measured). 7. Two of the following lines slope upwards and two downwards. B determining the four gradients sa which lines slope up and which down. (a) F( 1, 2), G(3, 8) (b) U( 3, 9), V(5, 5) (c) P(2, 12), Q(4, 2) (d) S(10, 4), T(2, 4). 8. Find the gradients of the two lines joining A( 5, 2) to B(3, 2) and W(1, 5) to Z(5, 3). What do our two answers tell ou about the line AB and the line XZ? 9. Two points have coordinates P(2, 1) and Q(6, b). The gradient m PQ = 2. Set up an equation and find the value of b. 10. Determine the values of p, q and r, given each of the following pairs of points along with the gradient joining each pair. (a) A(3, 1), B(5, p), m AB = 4. (b) L(4, 1), M(0, q), m LM = 2. (c) S( 1, 2), T(5, r), m ST = Draw a Cartesian diagram and plot the 4 points A( 2, 3), B(5, 0), C(6, 5) and D( 1, 2). (a) Show that m BC = m AD. (b) Show also that m AB = m DC. (c) What kind of quadrilateral does this prove ABCD is? (i.e. opposite pairs of sides are parallel). 12. In a similar wa, show that PQRS is also a parallelogram, where P(0, 1), Q(5, 1), R(8, 6) and S(3, 4). 13. It is known that IJKL is a parallelogram, where I(1, 3), J( 1, 1) and K( 2, 5). Find the gradients of the two lines IL and KL. (A sketch will help). 14. R is the point ( 6, 3), S(3, 6) and T( 2, 1). (a) Find the gradient m RT and m TS. (b) What does this prove about the 3 points R, T and S? (c) Plot the three points on a coordinate diagram and check this out. Collinear Points If three or more points lie on a straight line, the are said in mathematics to be collinear. To prove three points, A, B and C are collinear :- C Find m AB and m BC and show the are the same. State that AB and BC have a common point B. This proves the 3 points are collinear. 15. Prove that the three points U( 4, 6), V(2, 3) and W(10, 1) are collinear. (Hint :- Find m UV and m VW first and then...) 16. Prove that each set of points below are collinear :- (a) F(3, 1), G(5, 3) and H(8, 6). (b) R( 1, 8), S(5, 5) and T(9, 3). (c) D( 5, 15), E( 3, 12) and F(5, 0). 17. Three points, F, G and H are collinear. F is the point with coordinates, ( 3, 7), G is the point ( 1, 3) and H has coordinates (2, p). Determine the value of p. 18. L(8, 2), M(2, t) and N( 1, 1) are collinear. Find the value of t. 19. From his cabin, Ted walks to point R, 50 metres East and 35 metres West of the cabin. From R, he travels on a straight line to S, then on a straight line from S to T, as shown below. O R(50, 35) Had Ted walked on a straight line from R to T? A S(110, 75) B T(155, 105) Chapter H2 this is page 210 Equation of a Line

3 Angle between a Line and the -ais Since m AB = , we can see from the diagram that :- tanθ = = m AB (since tan = opp adj ) A θ θ 2 1 this is the positive direction of the -ais B 2 1 This means that if ou input the gradient of a line into our calculator and press Inverse - Tangent, ou will obtain the angle between the line and the positive direction of the -ais. Eample 1 :- Calculate the size of the angle the line through A(2, 1) and B(8, 3) makes with the positive direction of the -ais. m AB = = 3 ( 1) 8 2 = 4 6 = => θ = Inv-Tan ( ) = 33 7 Eample 2 :- A line makes an angle of 76 with the -ais. Calculate the gradient of the line. The angle the line makes is 76. => gradient = tan 76. => gradient = 4 0 Eercise H Cop and complete the following to determine the angle the line through P(1, 3) and Q(8, 5) makes with the -ais :- m AB = = => θ = = Calculate the size of the angle the line through each pair of points below makes with the -ais :- (a) through E(8, 2) and F(10, 7) (b) through M(0, 3) and N(5, 5) (c) through U( 1, 4) and T(5, 1) (d) through C(0, 5) and D(7, 3). = The line through A(1, 8) and B(7, 5) slopes downwards and has a negative gradient. Calculate the size of the angle this line makes with the positive direction of the -ais. (i.e. - the obtuse angle). 4. Calculate the size of the obtuse angle the line through each pair of points below makes with the -ais :- (a) through C(10, 2) and D(8, 7) (b) through M(0, 3) and N(5, 5). 5. A line makes an angle of 63 4 with the positive direction of the -ais. Calculate its gradient. (to 1 decimal place). 6. Calculate the gradient of the line which makes an angle of 71 5 with the positive direction of the -ais, to 1 decimal place. 7. Calculate the gradient of the line which makes an angle of 14 with the positive direction of the -ais, to 1 decimal place. 8. Calculate the gradient of the line which makes an angle of with the positive direction of the -ais. (Careful!) 63 4 Chapter H2 this is page 211 Equation of a Line

4 The Equation of a Straight Line - Revision In Book IC1, we found that the equation of (almost) an line takes the form : = m + c Where m represents the gradient of the line and c represents the -intercept. (where it cuts the -ais). gradient -intercept Eamples : (a) = has gradient 3 and -intercept 4. (b) A line with -intercept 2 and gradient 3 4 has equation = (c) B m AB = 3 6 = 1 2 -intercept is 1 => Equation of line AB is = A Eercise H Write down the gradient and -intercept in each of these equations : (a) (b) (a) = (b) = 6 4 (c) = (d) = (e) = 1 7 (f) = (c) (d) (g) = 3 (h) = Write down the equation of each of these lines : (a) m = 2, and the -intercept is 1. (b) m = 3, and the -intercept is 5. (c) gradient is 1, and it passes through (0, 4). 2 (d) m = 6 and line passes through the origin. 3. Line PQ cuts the -ais at the point (0, 4) and is parallel to a line with equation = 3 5. (a) Write down the gradient of the line PQ. (b) Write the equation of this line PQ. 4. For each of the lines shown at the top of the net column, (i) (ii) (iii) calculate the gradient, write down the -intercept, write the equation of the line. 5. The line shown has equation = m + c. (a) Write down the value of c. (b) Calculate the gradient. (c) Write down the equation of the line. 6. Use the same technique to determine the equation of this line :- (4, 6) (0, 2) (0, 1) (9, 4) Chapter H2 this is page 212 Equation of a Line

5 7. Write down the equation of each line below : (a) (b) (5, 7) 11. Match each of the following equations with their corresponding graphs shown below : (a) = 4 (b) = 3 2 (c) = 3 (d) = 4 (0, 2) (0, 6) (9, 3) (e) = (f) = 3. (c) (6, 4) (d) (0, 4) A B (0, 3) (6, 3) C D 8. A line passes through R(0, 2) and S(3, 5). (a) Show the line RS on a Cartesian diagram. (b) Calculate the gradient of this line. (0, 2) (0, 4) (c) Write down its -intercept. (d) Write down the equation of the line RS. E F 9. Find the equation of the line shown opposite (0, 3) Find the equation of the line passing through each set of points below :- (a) A(0, 3), B(2, 9) (b) U(0, 1), V(4, 1) Write down the equations of these lines :- (a) (b) (6, 6) (6, 4) (c) M( 2, 4), N(0, 4) (d) Z(0, 10), W(3, 5). 13. Show that the line through the pair of points A(0, 2) and B(4, 8) and the line through C(0, 5) and D(4, 1) are parallel to each other. (Remember : parallel lines have equal gradients) (0, 3) (0, 2) 14. Show that the line through the pair of points (2, 7) and (10, 3) and the line through ( 3, 4), and (15, 2) are NOT parallel to each other. (c) (d) 15. Write down the equation of the line : (0, 3) (3, 0) (10, 0) (a) which goes through the point (0, 2) and is parallel to the line = 3 5. (0, 2) (b) parallel to the line = and passing through the point (0, 4). Chapter H2 this is page 213 Equation of a Line

6 The Distance Formula In book IC1, ou learned that, given the coordinates of an two points P( 1, 1 ) and Q( 2, 2 ), ou could construct a right angled triangle, (as shown), and use Pthagoras Theorem to calculate the distance from P to Q. At Higher level, we have a more formal formula we use to calculate the distance between the two points. PQ 2 = PR 2 + RQ 2 => PQ 2 = ( 2 1 ) 2 + ( 2 1 ) 2 P( 1, 1 ) ( 2 1 ) Q( 2, 2 ) ( 2 1 ) R => PQ = ( 2 1 ) 2 + ( 2 1 ) 2 This is called the distance formula. Eample :- Calculate the length of the line AB, given that A( 2, 3) and B(6, 4). AB = ( 2 1 ) 2 + ( 2 1 ) 2 B(6, 4) => AB = (6 ( 2)) 2 + (4 ( 3)) 2 => AB = = 113 = 10 6 A( 2, 3) Eercise H Calculate the length of the line RS where R(3, 1) and S(5, 7). Cop and complete :- RS = ( 2 1 ) 2 + ( 2 1 ) 2 RS = (5...) 2 + (... ( 1)) 2 RS = =... = Calculate the lengths of the lines joining each pair of points listed below :- (a) D(2, 1) and E(6, 5) (b) P(5, 0) and Q(1, 8) (c) F( 1, 2) and Q(5, 3) (d) U( 2, 5) and V(1, 3). 3. For the 4 points J( 1, 4), K(5, 3), L(4, 3) and M(0, 5), determine which is longer - the line JL or the line KM. 4. Plot the three points C(2, 7), D(5, 3) and E( 2, 4) on a coordinate diagram. (a) Calculate the length of the line CD. (b) Calculate the length of the line CE. (c) What kind of triangle must CDE be? 5. Show that the three points A(5, 1), B( 2, 2) and C( 3, 5) could lie on the circumference of a circle with centre P(1, 2). 6. Draw a new set of aes and plot the following three points :- R( 2, 5), S(6, 1) and T(3, 5). (a) Calculate the lengths of the three sides RS, ST and RT. (b) Use the Converse of Pthagoras Theorem to prove that RST is a right angled triangle. Chapter H2 this is page 214 Equation of a Line

7 The General Equation of a Line A + B + C = 0 (Revisited) From Book IC1, ou learned the equation of an line can be written in the form A + B + C = 0, where A, B and C are Integers, (A, B 0 at the same time). A + B + C = 0 is called the General Equation of a line. This General Equation can be re-arranged so that the gradient and -intercept are easier to obtain. Eample 1 :- Find the gradient and intercept of the line = = 0 => 3 = 4 6 => = => m = 4 3, -intercept is (0, 2) Eample 2 :- Rearrange the line = into the form A + B + C = 0. = multipl both sides b 6 => 6 = => 6 = 4 3 => = 0 Eercise H The equation of a line is given b : = 0. Cop and complete the calculation below to find the gradient and the -intercept = 0 => 2 = => =... m =... and the -intercept is For each of the lines below : find the gradient and the -intercept. sketch each line. (a) = 0 (b) = 0 (c) = 0 (d) = 0 (e) 3 3 = (f) = Rearrange each of the following into the form A + B + C = 0 (a) = 2 11 (b) 3 = (c) 2 = (d) = 3 (e) 3 = 3 (f) = 4. (g) The line with gradient 3 through (0, 2) (h) The line through (0, 3) and (5, 8) (i) The line through (0, 6) and (5, 4) (j) The horizontal line through ( 3, 4) 4. (a) Sketch the line = 0. (b) On the same diagram, sketch the line = 0 and write down the coordinates of the intersection of the lines. 5. Re-arrange each set of equations where necessar and use simultaneous equations techniques to solve each set : (a) + 3 = 0 (b) = = = 0 (c) 3 21 = 0 (d) = = = 0 (e) = 0 (f) + 2 = 2 + = = Given that (p, 1) lies on the line 3 2 = 1, find the value of p. 7. The coordinates of three of the vertices of parallelogram ABCD are A(2, 1), B(6, 2), and C(4, 4). Find the general equation of the diagonal DB. 8. The line = is the R ais of smmetr of S the kite PQRS which has vertices at P(2, 2), P Q Q(4, 2), and S(2, 4). SR has the general equation = 0. Find the coordinates of the verte R. Chapter H2 this is page 215 Equation of a Line

8 The Equation of a Straight Line in the form b = m( a) Consider the line with gradient m passing through the point P(a, b). (This time we do not know the -intercept can t use = m + c). Let Q(, ) be ANY other point on the line. b The gradient of the line PQ is m PQ =. a This equation can be rearranged to give :- b = m( a) ****** P(a, b) m Q(, ) This is the MOST IMPORTANT equation in this chapter. It allows ou to determine the equation of an line as long as ou know one point (a, b) and the gradient m of the line. Eample 1 : Find the equation of the line with m = 2 through the point (1, 5). b = m( a) 5 = 2( 1) 5 = 2 2 = Eample 2 : Find the equation of the line through the two points (3, 5) and (1, 7). m = = = 1 b = m( a) 7 = 1( 1) 7 = + 1 = + 8 Either point could be used. We chose (1,7) Eercise H Find the equation of the line with m = 3, through the point (6, 1). Cop and complete :- 2. Find the equation of each line given the gradient and a point on the line. (a) m = 5, (1, 2) (b) m = 2, (3, 2) (c) m = 4, ( 3, 4) (d) m = 1, ( 1, 2) (e) m = 2, (5, 8) (f) m = 4, (0, 3) (g) (0, 0), m = 1 (h) (0, 3), m = Find the equation of the line : b = m( a) 1 =...(...)... = = (a) through the point (4, 5) and parallel to the line = 5 1. (b) parallel to + = 10 through the origin. (c) through (3, 5) and parallel to = 3 1. (d) through ( 1, 5), parallel to + 2 = 0. (e) parallel to the line = 0 and passing through the point ( 3, 2). 4. Find the equation of the line through the points (2, 3) and (5, 9). Cop and complete. = Find the equation of the line which passes through each set of points : (a) (1, 6), (2, 8) (b) (2, 5), (4, 9) (c) (3, 3), (4, 4) (d) (2, 6), (6, 2) (e) ( 2, 1), (4, 7) (f) ( 5, 1), ( 1, 1). 6. The line AB passes through A(4, 2) and B(6, 3). Show that C( 28, 16) does not lie on the line. 7. A parallelogram JKLM has vertices at J(3, 5), K(1, 2), L( 3, 5) and M( 1, 2). Find the equations of the two diagonals of the parallelogram. 8. The line CD passes through the points C(1, 3) and D(d, 5). Given that the line has equation :- m = = = b = m( a)... =...(...)... = = 1 ( 1), find the value of d. 4 Chapter H2 this is page 216 Equation of a Line

9 9. A line passing through P( 2, 2) and T(6, t), has its equation 2 6 = 0. Find the value of t. 10. P has coordinates (3, 7). The lines, 6 2 = 0 and = 0, intersect at the point Q. Find the equation of the line PQ. 12. An isosceles triangle STV has its base ST along the -ais with SV and VT having equal sides. The triangle has an ais of smmetr with equation = 7. S is the point ( 1, 0), and the line SV has equation = 0. S V T 11. Triangle ABC has vertices A( 6, 0) and B(8, 0). C (a) Find the coordinates of T and V. (b) Find the equation of the line VT. (c) Find the sizes of angles STV and SVT. A The line OC has equation = 7. The line AC has equation = + 6 (a) Determine the coordinates of C. B 13. An equilateral triangle OPQ has vertices at the origin and at Q(10, 0). (a) Show that the verte at P is (5, 5 3). O P Q (b) Find the equation of the line CB. (b) Find the equation of the line PQ. Perpendicular Lines Two lines are said to be perpendicular if the intersect at right angles. The figure shows the point P(a, b) being given a rotation of +90 about O. From the diagram, we see that :- P(a, b) Q( b, a). Let the gradient of OP = m 1 and the gradient of OQ = m 2. Then m 1 = b a and m 2 = a b. => m 1 m 2 = b a a b = 1, a 0, b 0. If OQ is perpendicular to OP, then m 1 m 2 = 1 Q( b, a) a m 2 b m 1 a +90 P(a, b) b The converse is also true :- If m 1 m 2 = 1, then OQ is perpendicular to OP. NOTE :- this is not true for lines parallel to the and the -aes Eample 1 :- Find the equation of the line through the point ( 3, 1) which is perpendicular to the line with equation : = = 0 rearranges to = So the gradient of the given line is 1 2 and the gradient of the perp. line is 2. Using b = m( a) gives... 1 = 2( + 3) i.e. = Eample 2 :- Prove that the line with equation = 0 is perpendicular to the line = = 0 rearranges to = => m 1 = 1 3 For = 3 1, m 2 = 3. As m 1 m 2 = = 1 the lines are perpendicular Chapter H2 this is page 217 Equation of a Line

10 Eercise H Which of the following pairs of numbers could be gradients of perpendicular lines? (a) 5 and 1 5 (b) 4 and (c) 5 and 5 2 (d) 5 4 and 4 5. (e) 0 3 and 3 (f) and 8 3 (g) and 2 3 (h) 1 6 and Write down the gradients of the lines perpendicular to the lines with gradients :- (a) 1 2 (b) 3 (c) 7 (d) 7 2 (e) 1 (f) 5 9 Vertical and Horizontal Lines B now, ou should know that lines running parallel to the -ais have equation = (a number) and line running parallel to the -ais have equation = (a number) 3 The red line has equation = 5 (m = 0). The green line has equation = 2 (m = 0). The brown line has equation = 6 (m =?) The gre line has equation = 3 (m =?) (g) 5 5 (h) (i) 0. careful! 3. Find the equation of the line which is perpendicular to the line with equation = 5, but which passes through the point (0, 4). 4. Find the equation of the line which passes through the point A( 2, 3), and which is perpendicular to the the line = Decide which of the following four pairs of lines are perpendicular to each other. (a) = and = (b) = and = (c) 2 + = 5 and 2 = 3 (d) 3 2 = 0 and = Determine the equation of the line through the point (5, 2), perpendicular to the line with equation = Find the equations of the lines through the origin, perpendicular to these lines :- (a) 5 = 1 (b) = Find the equation of the line perpendicular to the line = 0, which passes through the point (2, 4). 9. The line through (2, 1), which is perpendicular to the line p 7 = 0 is = q. Determine the value of p and then q. We sa the gradient of an line running parallel to the -ais is undefined - (it is infinitel large). 10. Write down the equation of the line parallel to :- (a) the -ais, passing through the point (3, 1). (b) the -ais, passing through the point ( 2, 4). (c) the -ais, passing through the point ( 1, 5). 11. Find the equation of the line perpendicular to :- (a) the line = 5, through the point (7, 2). (b) the line = 2, through the point (4, 6). (c) the line = 0, through the point ( 4, 8). 12. P is the point ( 4, 3), Q( 6, 3) and R(2, 1). Find the equation of the line through :- (a) P, parallel to QR. (b) P, perpendicular to QR. (c) Q, parallel to PR. (d) Q, perpendicular to PR. (e) R, parallel to PQ. 13. K is (1, 2), L(3, 4), M( 5, 1) and N(1, k). Find k, given that KL is perpendicular to MN. 14. P is (8, 1), Q( 7, 3), R( 6, 6) and S( 3, 5). Prove that line PR is perpendicular to line QS. 15. Prove that the lines with equations p + q + r = 0 and q p + s = 0, (p, q 0), are perpendicular to each other. 16. AB makes an angle of 63 5 with the positive direction of the -ais. RT is perpendicular to AB. Find m RT. Chapter H2 this is page 218 Equation of a Line

11 Medians, Altitudes and Perpendicular Bisectors Median :- (ou might like to tr Eercise H2 8, questions 1-4 after this). Given the coordinates of D, E and F in triangle DEF, the line ER is known as the median from the point E to the line DF. (There are two more medians). R is the mid-point of DF. E(..,..) F(..,..) Rule for finding the Equation of a Median (in this case ER) Find R, the mid-point of DF, using 1 + 2, Use this point and point E to find the gradient of ER - m ER Use either point E or point R along with m ER to find the equation b using [ b = m( a)]. D(..,..) R This is the median from E to the mid-point of AB. Altitudes :- (ou might like to tr Eercise H2 8, questions 5-7 after this). Given the coordinates of A, B and C in triangle ABC, the line CT is known as the altitude from the point C to the line AB. (There are 2 other altitudes from A and from B). CT meets AB at right angles. C(..,..) Rule for finding the Equation of an Altitude (in this case CT) Find the gradient of AB. ( m AB ) Find the perpendicular gradient. ( m CT ) Use point C along with m CT to find the equation b using [ b = m( a)]. A(..,..) B(..,..) T Note :- as et, we do NOT know the coordinates of T. Shown here is the altitude from C to AB. Perpendicular Bisector :- Given the coordinates of J, K and L in triangle JKL, the Perpendicular Bisector of KL is shown in the diagram. This Perpendicular Bisector cuts KL at its mid-point and is also at right angles to KL. (ou might like to tr Eercise H2 8, questions 8-11 after this). (There are two more perpendicular bisectors in this triangle). K(..,..) S L(..,..) Rule for finding the Equation of a Perpendicular Bisector (e.g. of KL) Find the coordinates of S, the mid-point of KL using 1 + 2, Find the gradient of line KL - m KL. Find the perpendicular gradient. ( m perp ). Use m perp and point S to find the equation using [ b = m( a)]. J(..,..) This is the perpendicular bisector of the line KL Note :- in a triangle, a Perpendicular Bisector need not go through an of the vertices. Chapter H2 this is page 219 Equation of a Line

12 ` Eample 1 :- Find the equation of the perpendicular bisector of PQ if P is (3, 2) and Q is ( 5, 4). Solution :- Make a sketch! Q( 5, 4) perpendicular bisector Midpoint of PQ = ( 1, 1) m PQ = 4 ( 2) 5 3 = 6 8 = 3 4 => m perp = 4 3 Equation of perpendicular bisector is 1 = 4 ( + 1) 3 P(3, 2) => = (or in an equivalent form). Eample 2 :- In the diagram, triangle ABC has vertices as shown. (a) Find the equation of the median from A to BC. (b) Find the equation of the altitude from B to AC. (c) Find the coordinates of W where the two lines cross. Solution :- A( 7, 5) B( 1, 10) 2 E W D C(8, ) (a) Midpoint of BC = E(4 1, 6 1 ) (b) m 4 4 AC = = m AE = = 1 m perp = 2 Using point A, the equation of Using point B, the equation of the median is + 5 = 1( + 7) the altitude is 10 = 2( 1 2 ) => = + 2 => = (c) Now solve :- = + 2 = => = 2 => + 2 = 11 simultaneous equations => = 3, = 5 W(3,5) Eercise H2 8 Where there is no diagram, it is advisable to make a sketch. 1. A, B and C are the points (6, 3), ( 2, 5) and ( 3, 1) respectivel. Find the equation of the median from C to AB. 2. Find the equation of the median, from P, of triangle PQR where the coordinates of P, Q and R are ( 2, 3), ( 3,,4) and (5, 2) respectivel. 3. The diagram shows a triangle KLM with vertices K(6, 5), L(8, 3) and M(0 1). Find the equation of the median through K. K(6, 5) M(0, 1) L(8, 3) 4. The vertices of a triangle are J( 6, 1), T(4, 5) and G(8, 3). (a) Find the equation of the median from J to TG. (b) Find the equation of the median from T to JG. (c) Find the coordinates of the point of intersection of the two medians. (d) Prove that the three medians of triangle JTG are concurrent (i.e. the all meet at the same point), b showing that the point of intersection lies on the third median from G to JT. Chapter H2 this is page 220 Equation of a Line

13 5. P( 6, 5), Q( 4, 2) and R(2, 1) are the vertices of triangle PQR. P( 6, 5) Q( 4, 2) A R(2, 1) Find the equation of PA, the altitude from P. 6. Triangle ABC has vertices A( 1, 1), B(2, 1) and C( 6, 2). Find the equation of the altitude of triangle ABC, drawn from A. 11. F is the point ( 3, 1), K is (5, 5) and Q is (6, 2) in triangle FKQ. (a) Find the equation of the perpendicular bisector of the line FK. (b) Find the equation of the perpendicular bisector of the line FQ. (c) Find the coordinates of the point of intersection of the two perpendicular bisectors. (d) Prove that the three perpendicular bisectors of triangle FKQ are concurrent b showing that the point of intersection lies on the perpendicular bisector of KQ. 12. Triangle ABC has vertices A( 6, 3), B(2, 7) and C(5, 9). C 7. H is the point (6, 6), R is ( 3, 0) and Y is (0, 3) in triangle HRY. (a) Find the equation of the altitude from H to RY. (b) Find the equation of the altitude from R to HY. (c) Find the coordinates of the point of intersection of the two altitudes. (d) Prove that the three altitudes of triangle HRY are concurrent b showing that the point of intersection lies on the third altitude. 8. D and E have coordinates (1, 0) and ( 5, 4) respectivel. (a) Write down the coordinates of the mid-point of DE and the gradient of DE. (b) Find the equation of the perpendicular bisector of DE. A (a) Find the equation of the median from C to AB. (b) Prove that this median is at right angles to side AB. (c) What kind of triangle must ABC be? 13. Triangle PAG has vertices P( 4, 10), A(10, 3) and G(0, 10) as shown. P B T A 9. Find the equation of the perpendicular bisector of the line joining M(2, 1) and N(8, 3). M 10. N( 2, 1), T( 4, 3) and L(8, 1) are the vertices of triangle NTL. Find the equation of the perpendicular bisector of the line NT and the line TL. G Find the point of intersection of the median AM and the altitude GT. Chapter H2 this is page 221 Equation of a Line

14 14. Triangle NUQ has vertices N(3, 3), U(13, 3) and Q(9, 7). Q 17. Triangle TVP has vertices T( 4, 8), V( 1, 2) and P( 7, 2). T (a) Write down the equation of the perpendicular bisector of NU. (b) Find the equation of the perpendicular bisector of NQ. (c) Find the coordinates of the point of intersection of these two lines. 15. Triangle DRP has coordinates D(1, 1), R(3, 15) and P(11, 5). The equation of the median from D to RP is (a) (b) (c) N D R 2 = 3 1. Find the equation of the median through the point R. Determine the coordinates of the point of intersection of the two medians. Write down the equation of the altitude through P. 16. Triangle EBY has vertices E( 3, 3), B( 1, 1) and Y(7, 3). (a) Prove that triangle EBY is right angled. (b) The median from E to BY meets the median from B to EY at point K. Find the coordinates of K. 2 = 3 1 P U P (a) Show that triangle TVP is isosceles. (b) The altitudes TA and VB intersect at W, where A and B lie on VP and PT respectivel. Find the coordinates of W. (c) Hence show that W lies one quarter wa up AT. 18. A, B and C are the points (5, 3), ( 8, 7) and (1, 5) respectivel. (a) Find the equation of the altitude of triangle ABC from A to BC. (b) The line through C parallel to the -ais meets this altitude at H and the line through A parallel to the -ais meets BC at L. Show that LH is the perpendicular bisector of the line AC. 19. A triangle has vertices P(6, 3), Q( 4,1) and R(4, 7). Q (a) Find the equations of the medians PL and QM. (b) Find the coordinates of J, the point of intersection of PL and QM. (c) Find the equation of RJ, and show that RJ produced bisects PQ. V R P Chapter H2 this is page 222 Equation of a Line

15 Remember Remember...? Topic in a Nutshell 1. The line joining the two points J(1, 3) and K(3, k) has a gradient of 4. Determine the value of k. 2. Prove that the points G( 30, 10), H(10, 10) and I(90, 50) are collinear. 3. Calculate the size of the angle the line joining each pair of points below makes with the positive direction of the -ais. (a) S(1, 0), T(6, 5) (b) F( 3, 1), G(1, 3) 4. Calculate the gradient of each line below: (a) (b) Calculate the length of the line PT where P( 3, 2) and T(1, 1). 7. The general equation of the line FG is given as = 0. Find the general equation of the line which is parallel to FG and passes through the origin. 8. Find the equation of the line : (a) where m = 2 through the point (7, 2). (b) parallel to + 2 = 3, through ( 3, 4). 9. A quadrilateral has vertices at B(0, 2), C(5, 1) D(6, 4) and E(1, 3). (a) Find the equation of the diagonal BD. (b) Find the equation of the diagonal CE. (c) Find the point of intersection of the two diagonals Calculate the W( 7, 3) circumference of this circle, where WB is the diameter. B( 1, 5) 10. Write down the gradient of a line perpendicular to a line with gradient :- (a) m = 2 (b) m = 0 25 (c) m = A line is drawn joining the two points S(4, 1) and P(12, 3). Find the equation of the line perpendicular to SP which passes through S. 12. A quadrilateral has vertices J( 5, 2), K( 1, 5), L(2, 1) and M( 2, 2). (a) Prove that JKLM is a square. (b) Find the point where the diagonals intersect. 13. Find the equation of the perpendicular bisector of the line joining T(0, 2) and V(6, 4). 14. Triangle ABC has vertices as shown. (a) Find the equation of the median, BM. (b) Find the equation of the perpendicular bisector of AC. 15. A triangle has vertices at F( 3, 4), G( 3, 4) and H(3, 1). Find the point of intersection of the perpendicular bisector of FG, and the altitude from G to side FH. 16. Triangle DGK has vertices at D(0, 6), G( 2, 0) and K(8, 2). (a) Find the equation of the altitude through G. (b) Find the equation of the median through D. (c) Determine the point of intersection of these two lines. 17. Three of the vertices of square CDEF are C(2, 4), D(3, 0) and E(7, 1). (a) Find the equation of the sides CF and EF. (b) Now determine the coordinates of F. B(5, 5) M A(0, 2) C(8, 2) Chapter H2 this is page 223 Equation of a Line

16 And so ends the Credit / Intermediate 2 Course. Best of luck in our EXAMS. from all at TEEJAY. Chapter H2 this is page 224 Equation of a Line