( ) Straight line graphs, Mixed Exercise 5. 2 b The equation of the line is: 1 a Gradient m= 5. The equation of the line is: y y = m x x = 12.

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Baldric Jackson
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1 Stright line grphs, Mied Eercise Grdient m ( y ),,, The eqution of the line is: y m( ) ( ) + y + Sustitute (k, ) into y + k + k k Multiply ech side y : k k The grdient of AB is: y y So: ( k ) 8 k k 8 k k 8 k Multiply ech side y (8 k): k ( 8 k) Multiply ech term y : k 8 k k 8 k k k. So A nd B hve coordintes (, ) nd (8, ). The eqution of the line is: y y y 8 Multiply ech side y : ( ) y + The eqution of L is: y m ( ) y + The eqution of L is: y ( ) 8 ( ) 8 + y + Solve y + nd y + simultneously Sustitute 9 into y + : y 9 + The lines intersect t C(9, ). Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free.

2 The eqution of l is: y y y y Multiply ech side y : ( ) y y ( ) y Solve + y nd y simultneously. Sustitute: + ( ) Sustitute into y : y 9 The coordintes of C re (, ). (, y) (, ), (, y) ( 9, 9) The eqution of L is: y y y 9 9 Multiply ech side y : ( ) Multiply ech term y : ( ) y + 9 y+ 9 The eqution of L is y + 9. (, y) (, ), (, y) (, ) The eqution of l is: y y y Multiply ech side y : Note: + + The eqution of l is: y m ( 9) 9 y + y + 9 A(, ), B ( +, + ) The grdient of the line through A nd B is: y y Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free.

3 Rtionlising the denomintor: (+ ) ( ) + (+ ) ( ) The eqution of the line is: y + c Sustituting nd y into y + c: + c c The eqution of line l is: y + Line l meets the -is when y. When y,. C is the point (, ). 8 A(, ), B(, 8) The grdient of AB is: y y 8 The grdient of line perpendiculr to AB is: The eqution of p is: y m ( ) 8 ( ) 8 + y + Sustitute in the eqution for AB: y () + The coordintes of C re (, ). 9 The line psses through A(, ) nd is perpendiculr to l: y. y y The grdient of y is. The grdient of line perpendiculr to y is. The eqution of the line m is: y m ( ) ( ) y + Or, since A is y-intercept, the eqution cn e written once the +. grdient is known i.e y To find P, solve y + nd y simultneously. Sustitute: ( + ) + Sustitute into y + : y + + The lines intersect t P(, ), s required. c A line prllel to the line m hs grdient. The eqution of the line n is: y m ( ) ( ) y + Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free.

4 9 c To find Q solve y nd y + simultneously. Sustitute: + + Sustitute into y + : y + + The lines intersect t Q(, ). A(, ) nd B(, ) The grdient of the line through A nd B is: y y ( ) 9 The eqution of the line through A nd B is: y + c Sustituting nd y into y + c: () + c so c As in Q9, the point A is the y-intercept so the eqution cn e written once the grdient hs een clculted. l: y l: + y 8 To find point D, solve simultneously y sustituting l into l. + 8 when, + y 8, y D is the point (, ). The se of the tringle AC is units. The height of the tringle is units. Are ACD is units A(, ) nd B(, ) The eqution of l through A nd B is: y y y Multiply ech side y : ( ) Note: + y + + y The eqution of l through C(, ) with grdient is: y m ( ) ( ) ( ) + + y + Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free.

5 A(, ), B(, ) nd C(k, ) The grdient of AB is: y y ( ) ( ) 8 Since ABC is right ngle the grdient of BC is: y y So k k Multiply ech side y (k ): ( k ) k + k k c The eqution of the line pssing through B nd C is: y y y Multiply ech side y : ( ) + y + + y + d Rememer ngle B is right ngle. Use the digrm or the distnce formul to find the lengths AB nd BC. AB BC + Are of ABC 8 units The eqution of the line through (, ) nd (, ) is: y y y y ( ) ( ) + Multiply ech side y : ( ) ( + ) ( Note: nd ) Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free.

For the coordintes of A, sustitute y : + 8 8 8 8 8 The coordintes of A re (, ). For the coordintes of B, sustitute : + y 8 y 8 y 8 y 8 The coordintes of B re 8 The re of OAB is: 8 8,.

6 For the coordintes of A, sustitute y : The coordintes of A re (, ). For the coordintes of B, sustitute : + y 8 y 8 y 8 y 8 The coordintes of B re 8 The re of OAB is: 8 8,. Solve y + nd y simultneously. Sustitute: ( ) Now sustitute into y : y ( ) 8,. The coordintes of A re Rerrnge l: y + into the form y m + c: y y l hs grdient nd it meets the coordinte es t (, ). l hs grdient nd it meets the y-is t (, ). l meets the -is when y. Sustitute y into the eqution: l meets the -is t (, ). c The grdient of l is. The grdient of line perpendiculr to l is. The eqution of this line is: y m ( ) y+ y Multiply ech term y : y The eqution of the line is y. Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free.

7 A(, ) nd B(, ) The grdient of the line l through A nd B is: y y 8 The eqution of l is: y + c Sustituting nd y into y + c: () + c c 8 y + c + y The grdient of the line l is, the y- intercept is. y c Solve + y nd y simultneously. + ( ) 8 When 8: y ( 8) y C is the point ( 8, ). d The grdient of OA is: y y d The grdient of OC is: y y ( 8). e Therefore the lines OA nd OC re perpendiculr. OA ( ) + ( ) (( 8) ) + ( ) 8 f Are of OAB 8 units (, ) nd (, ) The distnce d etween the points is: d ( ) + ( y y ) (( ) ) + ( ) 9 + For points (, ) nd (, ),. Sustitute into. Distnce c For points (, ) nd ( 9, ),. Sustitute into. Distnce Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free.

8 d For points (, ) nd (, ),. Sustitute into. Distnce (, y) is point on y, so its coordintes re (, ). The distnce etween A(, ) nd (, ) is: d ( ) + ( y y ) ( ( )) + ( ) d 9 + When : y ( ) The point is (, ) e ( + )( ) or 8 When, y ( ) When, y () 8 The points re B(, ) nd C(, ). c The grdient of the line y is, so the perpendiculr line hs grdient. Its eqution is: y + c When nd y : ( ) + c c y + d Solving y + simultneously: + nd y BC ( ) + ( y y ) ( ( )) + ( ( )) Distnce from A(, ) to (, ) is: ( ( )) + ( ) + ( ) Are of tringle is:.8 units 8 Grdient of the line of est fit is: y y 9. Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free. 8

9 8 C P +. Using the point (, ):.() + C.P c is the grdient, which is the increse in cron dioide emissions in millions of tonnes for every million tonnes of oil pollution. d The model is not vlid for smll vlues of P s negtive mount of cron dioide emissions is not possile. Chllenge (, ), B(, 8) nd C(, ) The eqution of AB is: y y y ( ) ( ) 8 ( ) ( ) y+ + y + + y y The grdient of AB. The grdient of line perpendiculr to AB. The eqution of the perpendiculr to AB through C(, ) is: y ( ( )) y y + 8 Point D is where the line nd the perpendiculr intersect. Solve the equtions y nd y + 8 simultneously. + 8 Multiply ech term y Now sustitute into y + 8: y () + 8 y D is the point (, ). Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free. 9

10 Chllenge AB ( ) + ( y y ) ( ( )) + (8 ( )) + CD ( ( )) + ( ) 8 + ( ) 8 Are of ABC 8 units A(, 8), B(9, 9) nd C(, ) The grdient of AB is: y y l is perpendiculr to AB, so its grdient is. It psses through C, so its eqution is: y + c () + c c The eqution of l is y +. The grdient of BC is: y y 9 9 l is perpendiculr to BC, so its grdient is. It psses through A, so its eqution is: y + c 8 () + c c 8 The eqution of l is y + 8. The grdient of AC is: y y 8 l is perpendiculr to BC, so its grdient is. It psses through B, so its eqution is: y + c 9 (9) + c c The eqution of l is y +. Solve l nd l simultneously y ( 8 9 ) + 9 Their point of intersection is ( 8 9, 9 ). Now solve l nd l simultneously y ( 8 ) Their point of intersection is ( 8, ). 9 9 Therefore, l, l nd l ll intersect t ( 8, ). 9 9 A(, ), B(, ) nd C(c, ) The grdient of AB is: y y l is perpendiculr to AB so its grdient is. Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free.

11 It psses through C so its eqution is: y + k where k is the y-intercept. At C, c nd y. c + k c k Now solve l nd l simultneously. c ( ) y The intersection of l nd l is the point, c. Therefore, l, l nd l ll intersect t, c. The eqution of line l is: c y + The grdient of BC is: y y c c l is perpendiculr to BC so its grdient is c. It psses through A, so its eqution is: c y + K where K is the y-intercept. At A,, y. c () + K K The eqution of line l is c y. l is the verticl line through (, ), so its eqution is. Solve l nd l simultneously. c y + c ( ) The intersection of l nd l is the point c ( ) (, ). Person Eduction Ltd. Copying permitted for purchsing institution only. This mteril is not copyright free.

R(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of

R(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of Higher Mthemtics Ojective Test Prctice ook The digrm shows sketch of prt of the grph of f ( ). The digrm shows sketch of the cuic f ( ). R(, 8) f ( ) f ( ) P(, ) Q(, ) S(, ) Wht re the domin nd rnge of