Coordinate Geometry Please choose a question to attempt from the following: 12345

STRAIGHT LINE : Question 1 Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). Go to full solution Go to Marker’s Comments Go to Main Menu Reveal answer only

STRAIGHT LINE : Question 1 Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). Go to full solution Go to Marker’s Comments Go to Main Menu Reveal answer only y = -5 / 3 x - 6

Markers Comments Begin Solution Continue Solution 3x – 5y = 4 3x - 4 = 5y 5y = 3x - 4 (5)(5) y = 3 / 5 x - 4 / 5 Using y = mx + c, gradient of line is 3 / 5 So required gradient = -5 / 3, ( m 1 m 2 = -1) We now have (a,b) = (-6,4) & m = -5 / 3. Using y – b = m(x – a) We get y – 4 = -5 / 3 (x – (-6)) y – 4 = -5 / 3 (x + 6) y – 4 = -5 / 3 x - 10 y = -5 / 3 x - 6 Question 1 Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4). Back to Home

3x – 5y = 4 3x - 4 = 5y 5y = 3x - 4 (5)(5) y = 3 / 5 x - 4 / 5 Using y = mx + c, gradient of line is 3 / 5 So required gradient = -5 / 3, ( m 1 m 2 = -1) We now have (a,b) = (-6,4) & m = -5 / 3. Using y – b = m(x – a) We get y – 4 = -5 / 3 (x – (-6)) y – 4 = -5 / 3 (x + 6) y – 4 = -5 / 3 x - 10 y = -5 / 3 x - 6 Markers Comments Back to Home An attempt must be made to put the original equation into the form y = mx + c to read off the gradient. State the gradient clearly. State the condition for perpendicular lines m 1 m 2 = -1. When finding m2 simply invert and change the sign on m1 m 1 = 3535 m 2 = -5 3 Use the y - b = m(x - a) form to obtain the equation of the line. Next Comment

Go to full solution Go to Marker’s Comments Go to Main Menu Reveal answer only STRAIGHT LINE : Question 2 Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3).

Go to full solution Go to Marker’s Comments Go to Main Menu Reveal answer only STRAIGHT LINE : Question 2 Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3). y = -2x + 7

Markers Comments Begin Solution Continue Solution Question 2 Back to Home 8x + 4y – 7 = 0 4y = -8x + 7 (4)(4) y = -2x + 7 / 4 y = -2x + 7 Using y = mx + c, gradient of line is -2 So required gradient = -2 as parallel lines have equal gradients. We now have (a,b) = (5,-3) & m = -2. Using y – b = m(x – a) We get y – (-3) = -2(x – 5) y + 3 = -2x + 10 Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3).

Markers Comments Back to Home Next Comment An attempt must be made to put the original equation into the form y = mx + c to read off the gradient. State the gradient clearly. State the condition for parallel lines m 1 = m 2 Use the y - b = m(x - a) form to obtain the equation of the line. 8x + 4y – 7 = 0 4y = -8x + 7 (4)(4) y = -2x + 7 / 4 y = -2x + 7 Using y = mx + c, gradient of line is -2 So required gradient = -2 as parallel lines have equal gradients. We now have (a,b) = (5,-3) & m = -2. Using y – b = m(x – a) We get y – (-3) = -2(x – 5) y + 3 = -2x + 10

Go to full solution Go to Marker’s Comments Go to Main Menu Reveal answer only STRAIGHT LINE : Question 3 In triangle ABC, A is (2,0), B is (8,0) and C is (7,3). (a) Find the gradients of AC and BC. (b) Hence find the size of ACB. X Y AB C

Go to full solution Go to Marker’s Comments Go to Main Menu Reveal answer only STRAIGHT LINE : Question 3 In triangle ABC, A is (2,0), B is (8,0) and C is (7,3). (a) Find the gradients of AC and BC. (b) Hence find the size of ACB. X Y AB C = 77.4° (b) m AC = 3 / 5 m BC = - 3 (a)

Markers Comments Begin Solution Continue Solution Question 3 Back to Home (a)Using the gradient formula: m AC = 3 – = 3 / 5 m BC = 3 – = - 3 In triangle ABC, A is (2,0), B is (8,0) and C is (7,3). (a)Find the gradients of AC and BC. (b) Hence find the size of ACB. (b) Using tan  = gradient If tan  = 3 / 5 then CAB = 31.0° If tan  = -3 then CBX = ( )° = o Hence : ACB = 180° – 31.0° – 71.6° = 77.4° so ABC = 71.6° X Y A B C

Markers Comments Back to Home Next Comment (a)Using the gradient formula: m AC = 3 – m BC = 3 – = - 3 (b) Using tan  = gradient = 3 / 5 If tan  = 3 / 5 then CAB = 31.0° then CBX = ( )° = o Hence : ACB = 180° – 31.0° – 71.6° = 77.4° If tan  = -3 If no diagram is given draw a neat labelled diagram. In calculating gradients state the gradient formula. Must use the result that the gradient of the line is equal to the tangent of the angle the line makes with the positive direction of the x-axis. Not given on the formula sheet. A B Ø ° m AB = tanØ ° Ø ° = tan -1 m AB so ABC = 71.6°

STRAIGHT LINE : Question 4 Go to full solution Go to Marker’s Comments Go to Main Menu Reveal answer only In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). X Y P(4,-5) Q(2,3) R(10,-1) Find (a) the equation of the line e, the median from R of triangle PQR. (b) the equation of the line f, the perpendicular bisector of QR. (c) The coordinates of the point of intersection of lines e & f.

STRAIGHT LINE : Question 4 Go to full solution Go to Marker’s Comments Go to Main Menu Reveal answer only In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). X Y P(4,-5) Q(2,3) R(10,-1) Find (a) the equation of the line e, the median from R of triangle PQR. (b) the equation of the line f, the perpendicular bisector of QR. (c) The coordinates of the point of intersection of lines e & f. y = -1(a) y = 2x – 11 (b) (5,-1) (c)

Markers Comments Begin Solution Continue Solution Question 4 (a) Back to Home In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). X Y P(4,-5) Q(2,3) R(10,-1) Find (a) the equation of the line e, the median from R of triangle PQR. (a)Midpoint of PQ is (3,-1): let’s call this S Using the gradient formula m = y 2 – y 1 x 2 – x 1 m SR = -1 – (-1) Since it passes through (3,-1) equation of e is y = -1 = 0(ie line is horizontal) Solution to 4 (b)

Markers Comments Begin Solution Continue Solution Question 4 (b) Back to Home (b) the equation of the line f, the perpendicular bisector of QR. In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Find X Y P(4,-5) Q(2,3) R(10,-1) (b) Midpoint of QR is (6,1) m QR = 3 – (-1) = 4 / -8 = - 1 / 2 required gradient = 2 (m 1 m 2 = -1) Using y – b = m(x – a) with (a,b) = (6,1) & m = 2 we get y – 1 = 2(x – 6) so f is y = 2x – 11 Solution to 4 (c)

Markers Comments Begin Solution Continue Solution Question 4 (c) Back to Home In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Find (c) The coordinates of the point of intersection of lines e & f. X Y P(4,-5) Q(2,3) R(10,-1) (c) e & f meet when y = -1 & y = 2x -11 so 2x – 11 = -1 ie 2x = 10 ie x = 5 Point of intersection is (5,-1)

Markers Comments Back to Home Next Comment If no diagram is given draw a neat labelled diagram. Q P R y x median Perpendicular bisector (a)Midpoint of PQ is (3,-1): let’s call this S Using the gradient formula m = y 2 – y 1 x 2 – x 1 m SR = -1 – (-1) Since it passes through (3,-1) equation of e is y = -1 (ie line is horizontal) Comments for 4 (b) Sketch the median and the perpendicular bisector

Markers Comments Back to Home Next Comment Q P R y x (b) Midpoint of QR is (6,1) m QR = 3 – (-1) = 4 / -8 required gradient = 2 (m 1 m 2 = -1) Using y – b = m(x – a) with (a,b) = (6,1) & m = 2 we get y – 1 = 2(x – 6) so f is y = 2x – 11 = - 1 / 2 To find midpoint of QR (-1) 2, Look for special cases: Horizontal lines in the form y = k Vertical lines in the form x = k Comments for 4 (c)

Markers Comments Back to Home Next Comment (c) e & f meet when y = -1 & y = 2x -11 so 2x – 11 = -1 ie 2x = 10 ie x = 5 Point of intersection is (5,-1) y = -1 y = 2x - 11 To find the point of intersection of the two lines solve the two equations:

STRAIGHT LINE : Question 5 Go to full solution Go to Marker’s Comments Go to Main Menu Reveal answer only In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). Find (a) the equation of the altitude from vertex E. (b) the equation of the median from vertex F. (c) The point of intersection of the altitude and median. X Y G(2,-5) E(6,-3) F(12,-5)

STRAIGHT LINE : Question 5 Go to full solution Go to Marker’s Comments Go to Main Menu Reveal answer only In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). Find (a) the equation of the altitude from vertex E. (b) the equation of the median from vertex F. (c) The point of intersection of the altitude and median. X Y G(2,-5) E(6,-3) F(12,-5) x = 6 (a) x + 8y + 28 = 0 (b) (6,-4.25) (c)

Markers Comments Begin Solution Continue Solution Question 5(a) Back to Home In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). Find (a) the equation of the altitude from vertex E. X Y G(2,-5) E(6,-3) F(12,-5) (a)Using the gradient formula m FG = -5 – (-5) = 0 (ie line is horizontal so altitude is vertical) Altitude is vertical line through (6,-3) ie x = 6 Solution to 5 (b)

Markers Comments Begin Solution Continue Solution Question 5(b) Back to Home In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). Find X Y G(2,-5) E(6,-3) F(12,-5) (b) the equation of the median from vertex F. (b)Midpoint of EG is (4,-4)- let’s call this H m FH = -5 – (-4) = -1 / 8 Using y – b = m(x – a) with (a,b) = (4,-4) & m = -1 / 8 we get y – (-4) = -1 / 8 (x – 4) (X8) or 8y + 32 = -x + 4 Median is x + 8y + 28 = 0 Solution to 5 (c)

Markers Comments Begin Solution Continue Solution Question 5(c) Back to Home In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5). Find X Y G(2,-5) E(6,-3) F(12,-5) (c) The point of intersection of the altitude and median. (c) Lines meet when x = 6 & x + 8y + 28 = 0 put x =6 in 2 nd equation 8y + 34 = 0 ie 8y = -34 ie y = Point of intersection is (6,-4.25)

Markers Comments Back to Home Next Comment If no diagram is given draw a neat labelled diagram. Sketch the altitude and the median. y x F E G median altitude (a)Using the gradient formula m FG = -5 – (-5) = 0 (ie line is horizontal so altitude is vertical) Altitude is vertical line through (6,-3) ie x = 6 Comments for 5 (b)

Markers Comments Back to Home Next Comment y x F E G Comments for 5 (c) (b)Midpoint of EG is (4,-4)- call this H m FH = -5 – (-4) = -1 / 8 Using y – b = m(x – a) with (a,b) = (4,-4) & m = -1 / 8 we get y – (-4) = -1 / 8 (x – 4) (X8) or 8y + 32 = -x + 4 Median is x + 8y + 28 = 0 To find midpoint of EG (-5) 2, H Horizontal lines in the form y = k Vertical lines in the form x = k Look for special cases: