Short Questions (Question 10)

Posted on May 18, 2020 by Myhometuition

Question 10 (4 marks):
Diagram 10 shows the position of three campsites A, B and C at a part of a riverbank drawn on a Cartesian plane, such that A and B lie on the same straight riverbank.

Diagram 10

Sam wants to cross the river from campsite C to the opposite riverbank where the campsites A and B are located.
Find the shortest distance, in m, that he can take to cross the river. Give your answer correct to four decimal places.

Solution:

$\begin{array}{l} Let the shortest distance from campsite C \\ to opposite riverbank is C D . \\ Area of △ A B C \\ = \frac{1}{2} | \begin{array}{l} - 2 7 4 \\ 35 - 3 \end{array} \begin{array}{l} - 2 \\ 3 \end{array} | \\ = \frac{1}{2} | (- 2) (5) + (7) (- 3) + (4) (3) \\ - (3) (7) - (5) (4) - (- 3) (- 2) | \\ = \frac{1}{2} | - 66 | \\ = 33 {units}^{2} \\ Distance of A B \\ = \sqrt{{(- 2 - 7)}^{2} + {(3 - 5)}^{2}} \\ = \sqrt{85} m \\ Area of △ A B C = 33 \\ \frac{1}{2} (A B) (C D) = 33 \\ \frac{1}{2} (\sqrt{85}) (C D) = 33 \\ C D = \frac{33 (2)}{\sqrt{85}} \\ C D = 7.1587 (4 d .p .) \\ Thus, the shortest distance is 7 .1587 m . \end{array}$

Short Questions (Question 8 & 9)

Posted on May 18, 2020 by Myhometuition

Question 8 (3 marks):
The following information refers to the equation of two straight lines, AB and CD.

$\begin{array}{l} A B : y - 2 k x - 3 = 0 \\ C D : \frac{x}{3 h} + \frac{y}{4} = 1 \\ where h and k are constants . \end{array}$

Given the straight lines AB and CD are perpendicular to each other, express h in terms of k.

Solution:

$\begin{array}{l} A B : y - 2 k x - 3 = 0 \\ y = 2 k x + 3 \\ m_{A B} = 2 k \\ C D : \frac{x}{3 h} + \frac{y}{4} = 1 \\ m_{C D} = - \frac{4}{3 h} \\ m_{A B} \times m_{C D} = - 1 \\ 2 k \times (- \frac{4}{3 h}) = - 1 \\ - 8 k = - 3 h \\ h = \frac{8}{3} k \end{array}$

Question 9 (3 marks):
A straight line passes through P(3, 1) and Q(12, 7). The point R divides the line segment PQ such that 2PQ = 3RQ.
Find the coordinates of R.

Solution:

$\begin{array}{l} 2 P Q = 3 R Q \\ \frac{P Q}{R Q} = \frac{3}{2} \\ Point R \\ = (\frac{1 (12) + 2 (3)}{1 + 2}, \frac{1 (7) + 2 (1)}{1 + 2}) \\ = (\frac{18}{3}, \frac{9}{3}) \\ = (6, 3) \end{array}$

Long Questions (Question 10)

Posted on May 17, 2020 by Myhometuition

Question 10 (7 marks):
Solution by scale drawing is not accepted.
Diagram shows the locations of town M and town N drawn on a Cartesian plane.

PQ is a straight road such that the distance from town M and town N to any point on the road is always equal.
(a) Find the equation of PQ.

(b) Another straight road, ST with an equation y = 2x + 7 is to be built.
(i) A traffic light is to be installed at the crossroads of the two roads.
Find the coordinates of the traffic light.
(ii) Which of the two roads passes through town L

$(- \frac{4}{3}, 1) ?$

Solution:
(a)

$\begin{array}{l} T (x, y) is a point on P Q . \\ T M = T N \\ \sqrt{[x - {(- 4)}^{2}] + {[y - (- 1)]}^{2}} = \sqrt{{(x - 2)}^{2} + {(y - 1)}^{2}} \\ \sqrt{{(x + 4)}^{2} + {(y + 1)}^{2}} = \sqrt{{(x - 2)}^{2} + {(y - 1)}^{2}} \\ {(x + 4)}^{2} + {(y + 1)}^{2} = {(x - 2)}^{2} + {(y - 1)}^{2} \\ x^{2} + 8 x + 16 + y^{2} + 2 y + 1 \\ = x^{2} - 4 x + 4 + y^{2} - 2 y + 1 \\ 8 x + 2 y + 17 + 4 x + 2 y - 5 = 0 \\ 12 x + 4 y + 12 = 0 \\ 3 x + y + 3 = 0 \\ Equation of P Q : 3 x + y + 3 = 0 \end{array}$

(b)(i)

$\begin{array}{l} y = 2 x + 7 ............ (1) \\ 3 x + y + 3 = 0 ............ (2) \\ Substitute (1) into (2) : \\ 3 x + 2 x + 7 + 3 = 0 \\ 5 x = - 10 \\ x = - 2 \\ When x = - 2, \\ F r o m (1), \\ y = 2 (- 2) + 7 = 3 \\ Coordinates of traffic light = (- 2, 3) . \end{array}$

(b)(ii)

$\begin{array}{l} L (- \frac{4}{3}, 1) : x = - \frac{4}{3}, y = 1 \\ The equation of S T : y = 2 x + 7 \\ Left hand side: y = 1 \\ Right hand side: 2 (- \frac{4}{3}) + 7 = 4 \frac{1}{3} \\ Thus, the road y = 2 x + 7 does not \\ pass through L . \\ The equation of P Q : 3 x + y + 3 = 0 \\ Left hand side: \\ 3 x + y + 3 = 3 (- \frac{4}{3}) + 1 + 3 \\ = - 4 + 4 = 0 \\ Right hand side = 0 \\ Left hand side = Right hand side \\ Thus, the road 3 x + y + 3 = 0 \\ passes through L . \end{array}$

Long Questions (Question 9)

Posted on May 17, 2020 by Myhometuition

Question 9 (6 marks):
Solution by scale drawing is not accepted.
Diagram shows a triangle OCD.

Diagram

(a) Given the area of triangle OCD is 30 units², find the value of h.

(b) Point Q (2, 4) lies on the straight line CD.
(i) Find CQ : QD.
(ii) Point P moves such that PD = 2 PQ.
Find the equation of the locus P.

Solution:
(a)

$\begin{array}{l} Given Area of △ O C D = 30 \\ \frac{1}{2} | \begin{array}{l} 0 h - 6 \\ 0 2 8 \end{array} \begin{array}{l} 0 \\ 0 \end{array} | = 30 \\ | (0) (2) + (h) (8) + (- 6) (0) - (0) (h) - (2) (- 6) - (8) (0) | = 60 \\ | 0 + 8 h + 0 - 0 + 12 - 0 | = 60 \\ | 8 h + 12 | = 60 \\ 8 h + 12 = 60 \\ 8 h = 48 \\ h = 6 \\ or \\ 8 h + 12 = - 60 \\ 8 h = - 72 \\ h = - 9 (ignore) \end{array}$

(b)(i)

$\begin{array}{l} [\frac{6 (m) + (- 6) (n)}{m + n}, \frac{2 (m) + (8) (n)}{m + n}] = (2, 4) \\ \frac{6 m - 6 n}{m + n} = 2 \\ 6 m - 6 n = 2 m + 2 n \\ 4 m = 8 n \\ \frac{m}{n} = \frac{8}{4} \\ \frac{m}{n} = \frac{2}{1} \\ \frac{2 m + 8 n}{m + n} = 4 \\ 2 m + 8 n = 4 m + 4 n \\ 2 m = 4 n \\ \frac{m}{n} = \frac{4}{2} \\ \frac{m}{n} = \frac{2}{1} \\ Thus, C Q = Q D = 2 : 1 \end{array}$

(b)(ii)

$\begin{array}{l} P D = 2 P Q \\ \sqrt{{(x - 6)}^{2} + {(y - 2)}^{2}} = 2 \sqrt{{(x - 2)}^{2} + {(y - 4)}^{2}} \\ {(x - 6)}^{2} + {(y - 2)}^{2} = 4 [{(x - 2)}^{2} + {(y - 4)}^{2}] \\ x^{2} - 12 x + 36 + y^{2} - 4 y + 4 = 4 [x^{2} - 4 x + 4 + y^{2} - 8 y + 16] \\ x^{2} - 12 x + 36 + y^{2} - 4 y + 4 = 4 x^{2} - 16 x + 16 + 4 y^{2} - 32 y + 64 \\ The equation of locus P : \\ 3 x^{2} + 3 y^{2} - 4 x - 28 y + 40 = 0 \end{array}$

Long Questions (Question 7 & 8)

Posted on May 16, 2020 by Myhometuition

Question 7:

The diagram shows a straight line PQ which meets a straight line RS at the point Q. The point P lies on the y-axis.

(a) Write down the equation of RS in the intercept form.

(b) Given that 2RQ = QS, find the coordinates of Q.

(c) Given that PQ is perpendicular to RS, find the y-intercept of PQ.

Solution:

(a)

Equation of RS

$\begin{array}{l} \frac{x}{12} + \frac{y}{- 6} = 1 \\ \frac{x}{12} - \frac{y}{6} = 1 \end{array}$

(b)

$\begin{array}{l} Given 2 R Q = Q S \\ \frac{R Q}{Q S} = \frac{1}{2} \\ Lets coordinates of Q = (x, y) \\ (\frac{(0) (2) + (12) (1)}{1 + 2}, \frac{(- 6) (2) + (0) (1)}{1 + 2}) = (x, y) \\ x = \frac{12}{3} = 4 \\ y = - \frac{12}{3} = - 4 \\ Q = (4, - 4) \end{array}$

(c)

$\begin{array}{l} Gradient of R S, m_{R S} \\ = - (\frac{- 6}{12}) = \frac{1}{2} \\ m_{P Q} = - \frac{1}{m_{R S}} = - \frac{1}{\frac{1}{2}} = - 2 \end{array}$

Point Q = (4, –4), m = –2

Using y = mx+ c

–4 = –2 (4) + c

c = 4

y–intercept of PQ = 4

Question 8:

In the diagram, the equation of FMG is y = – 4. A point P moves such that its distance from E is always half of the distance of E from the straight line FG. Find

(a) The equation of the locus of P,

(b) The x-coordinate of the point of intersection of the locus and the x-axis.

Solution:

(a)

Gradient of the straight line FMG = 0

EM is perpendicular to FMG, so gradient of EM also = 0, equation of EM is x = 2

Thus, coordinates of point M = (2, –4).

Let coordinates of point P= (x, y).

Given PE = ½ EM

2PE = EM

2 [(x – 2)²+ (y – 4)²]^½ = [(2 – 2)² + (4 – (–4))²]^½

4 (x² – 4x + 4 + y² – 8y +16) = (0 + 64) → (square for both sides)

4x² – 16x + 16 + 4y² – 32y + 64 = 64

4x² + 4y² – 16x – 32y + 16 = 0

x² + y² – 4x – 8y + 4 = 0

(b)

x² + y² – 4x – 8y + 4 = 0

At x axis, y = 0.

x² + 0 – 4x – 8(0) + 4 = 0

x² – 4x+ 4 = 0

(x – 2) (x – 2) = 0

x = 2

The x-coordinate of the point of intersection of the locus and the x-axis is 2.

Long Questions (Question 6)

Posted on May 16, 2020 by Myhometuition

Question 7:

Solutions by scale drawing will not be accepted.

Diagram below shows a triangle OPQ. Point S lies on the line PQ.

(a) A point Y moves such that its distance from point S is always 5 uints.

Find the equation of the locus of Y.

(b) It is given that point P and point Q lie on the locus of Y .

Calculate

(i) the value of k,

(ii) the coordinates of Q.

(c) Hence, find the area, in uint², of triangle OPQ.

Solution:
(a)

$\begin{array}{l} The equation of the locus Y (x, y) is given by Y S = 5 units \\ \sqrt{{(x - 5)}^{2} + {(y - 3)}^{2}} = 5 \\ x^{2} - 10 x + 25 + y^{2} - 6 y + 9 = 25 \\ x^{2} + y^{2} - 10 x - 6 y + 9 = 0 \end{array}$

(b)(i)

Given P (2, k) lies on the locus of Y.

(2)² + (k)²– 10(2) – 6(k) + 9 = 0

4 + k²– 20 – 6k + 9 = 0

k² – 6k – 7 = 0

(k – 7) (k + 1) = 0

k = 7 or k = – 1
Based on the diagram, k = 7.

(b)(ii)

As P and Q lie on the locus of Y, S is the midpoint of PQ. P = (2, 7), S = (5, 3).

Let the coordinates of Q = (x, y),

$\begin{array}{l} (\frac{2 + x}{2}, \frac{7 + y}{2}) = (5, 3) \\ \frac{2 + x}{2} = 5 and \frac{7 + y}{2} = 3 \\ 2 + x = 10 and 7 + y = 6 \\ x = 8 and y = - 1 \end{array}$

Coordinates of point Q = (8, –1).

(c)

$\begin{array}{l} Area of △ O P Q \\ = \frac{1}{2} | \begin{array}{l} 0 8 2 \\ 0 - 17 \end{array} \begin{array}{l} 0 \\ 0 \end{array} | \\ = \frac{1}{2} | 0 + (8) (7) + 0 - 0 - (- 1) (2) - 0 | \\ = \frac{1}{2} | 58 | \\ = 29 {units}^{2} \end{array}$

Long Questions (Question 5)

Posted on May 16, 2020 by Myhometuition

Question 5:
Diagram below shows a quadrilateral ABCD. Point C lies on the y-axis.

The equation of a straight line AD is 2y = 5x – 21
(a) Find
(i) the equation of the straight line AB,
(ii) the coordinates of A,
(b) A point P moves such that its distance from point D is always 5 units.
Find the equation of the locus of P.

Solution:
(a)(i)

$\begin{array}{l} 2 y = 5 x - 21 \\ y = \frac{5}{2} x - \frac{21}{2} \\ m_{A D} = \frac{5}{2} \\ m_{A B} \times m_{A D} = - 1 \\ m_{A B} \times \frac{5}{2} = - 1 \\ m_{A B} = - \frac{2}{5} \\ Equation of A B \\ y - y_{1} = m_{A B} (x - x_{1}) \\ y + 1 = - \frac{2}{5} (x + 2) \\ 5 y + 5 = - 2 x - 4 \\ 5 y = - 2 x - 9 \end{array}$

(a)(ii)

$\begin{array}{l} 2 y = 5 x - 21 .......... (1) \\ 5 y = - 2 x - 9 .......... (2) \\ (1) \times 5 : 10 y = 25 x - 105 .......... (3) \\ (2) \times 2 : 10 y = - 4 x - 18 .......... (4) \\ (2) - (4) : 0 = 29 x - 87 \\ x = 3 \\ From (1), \\ 2 y = 15 - 21 \\ 2 y = - 6 \\ y = - 3 \\ A = (3, - 3) \end{array}$

(b)

$\begin{array}{l} y = 2, \\ 4 = 5 x - 21 \\ 5 x = 25 \\ x = 5 \\ Point D = (5, 2) \\ P D = 5 \\ \sqrt{{(x - 5)}^{2} + {(y - 2)}^{2}} = 5 \\ {(x - 5)}^{2} + {(y - 2)}^{2} = 25 \\ x^{2} - 10 x + 25 + (y^{2} - 4 y + 4) = 25 \\ x^{2} + y^{2} - 10 x - 4 y + 4 = 0 \end{array}$

Long Questions (Question 4)

Posted on April 21, 2020 by user

Question 4:
Solutions by scale drawing will not be accepted.
Diagram below shows a triangle PRS. Side PR intersects the y-axis at point Q.

(a) Given PQ : QR = 2 : 3, find
(i) The coordinates of P,
(ii) The equation of the straight line PS,
(iii) The area, in unit², of triangle PRS.
(b) Point M moves such that its distance from point R is always twice its distance from point S.
Find the equation of the locus M.

Solution:
(a)(i)

$\begin{array}{l} P = (\frac{2 (6) + 3 h}{2 + 3}, \frac{2 (12) + 3 k}{2 + 3}) \\ (0, 6) = (\frac{12 + 3 h}{5}, \frac{24 + 3 k}{5}) \\ \frac{12 + 3 h}{5} = 0 \\ 3 h = - 12 \\ h = - 4 \\ \frac{24 + 3 k}{5} = 6 \\ 3 k = 30 - 24 \\ k = 2 \\ P = (- 4, 2) \end{array}$

(a)(ii)

$\begin{array}{l} m_{P S} = \frac{2 - (- 6)}{- 4 - 2} \\ = - \frac{8}{6} \\ = - \frac{4}{3} \\ Equation of P S : \\ y - y_{1} = - \frac{4}{3} (x - 2) \\ y - (- 6) = - \frac{4}{3} x + \frac{8}{3} \\ 3 y + 18 = - 4 x + 8 \\ 3 y = - 4 x - 10 \end{array}$

(a)(iii)

$\begin{array}{l} Area of △ P R S \\ = \frac{1}{2} | \begin{array}{l} - 4 2 6 \\ 2 - 612 \end{array} \begin{array}{l} - 4 \\ 2 \end{array} | \\ = \frac{1}{2} | (24 + 24 + 12) - (4 - 36 - 48) | \\ = \frac{1}{2} | 60 - (- 80) | \\ = 70 {unit}^{2} \end{array}$

(b)

$\begin{array}{l} Let P = (x, y) \\ M R = 2 M S \\ \sqrt{{(x - 6)}^{2} + {(y - 12)}^{2}} = 2 \sqrt{{(x - 2)}^{2} + {(y + 6)}^{2}} \\ {(x - 6)}^{2} + {(y - 12)}^{2} = 4 [{(x - 2)}^{2} + {(y + 6)}^{2}] \\ x^{2} - 12 x + 36 + y^{2} - 24 y + 144 = 4 [x^{2} - 4 x + 4 + y^{2} + 12 y + 36] \\ x^{2} - 12 x + y^{2} - 24 y + 180 = 4 x^{2} - 16 x + 4 y^{2} + 48 y + 160 \\ 3 x^{2} + 3 y^{2} - 4 x + 72 y - 20 = 0 \end{array}$

Long Questions (Question 3)

Posted on April 21, 2020 by user

Question 3:

The diagram shows a triangle LMN where L is on the y-axis. The equation of the straight line LKN and MK are 2y – 3x + 6 = 0 and 3y + x– 13 = 0 respectively. Find

(a) the coordinates of K

(b) the ratio LK:KN

Solution:
(a)

2y – 3x + 6 = 0 ----(1)

3y + x – 13 = 0 ----(2)

x = 13 – 3y ----(3)

Substitute equation (3) into (1),

2y – 3 (13 – 3y) + 6 = 0

2y – 39 + 9y + 6 = 0

11y = 33

y = 3

Substitute y = 3 into equation (3),

x = 13 – 3 (3)

x = 4

Coordinates of K = (4, 3).

(b)

Given equation of LKN is 2y – 3x + 6 = 0

At y – axis, x = 0,

x coordinates of point L = 0.

$\begin{array}{l} Ratio L K : K N \\ Equating the x coordinates, \\ \frac{L K (10) + K N (0)}{L K + K N} = 4 \\ 10 L K = 4 L K + 4 K N \\ 6 L K = 4 K N \\ \frac{L K}{K N} = \frac{4}{6} \\ \frac{L K}{K N} = \frac{2}{3} \\ Ratio L K : K N = 2 : 3 \end{array}$

Long Questions (Question 2)

Posted on April 21, 2020 by user

Question 2:

In the diagram, PRS and QRT are straight lines. Given R is the midpoint of PS and
QR : RT = 1 : 3, Find

(a) the coordinates of R,

(b) the coordinates of T,

(c) the coordinates of the point of intersection between lines PQ and ST produced.

Solution:
(a)

Given R is the midpoint of PS.

$\begin{array}{l} R = (\frac{3 + 7}{2}, \frac{2 + 6}{2}) \\ R = (5, 4) \end{array}$

(b)

$\begin{array}{l} Q R : R T = 1 : 3 \\ Lets coordinates of T = (x, y) \\ (\frac{(1) (x) + (3) (4)}{1 + 3}, \frac{(1) (y) + (3) (5)}{1 + 3}) = (5, 4) \\ \frac{x + 12}{4} = 5 \\ x + 12 = 20 \\ x = 8 \\ \frac{y + 15}{4} = 4 \\ y + 15 = 16 \\ y = 1 \\ T = (8, 1) \end{array}$

(c)

$\begin{array}{l} Gradient of P Q = \frac{5 - 2}{4 - 3} = 3 \\ Equation of P Q, \\ y - 2 = 3 (x - 3) \\ y - 2 = 3 x - 9 \\ y = 3 x - 7 - - - - (1) \\ Gradient of S T = \frac{6 - 1}{7 - 8} = - 5 \\ Equation of S T, \\ y - 1 = - 5 (x - 8) \\ y - 1 = - 5 x + 40 \\ y = - 5 x + 41 - - - - (2) \end{array}$

Substitute (1) into (2),

3x – 7 = –5x + 41

8x = 48

x = 6

From (1),

y = 3(6) – 7 = 11

The coordinates of the point of intersection between lines PQ and ST = (6, 11).