Coordinate Geometry Past Paper Questions PDF

## Coordinate Geometry (Chapter 6) 141 **Step 1:** Rearrange each equation into the form $y = mx + c$. * $4x + 3y = 10$ and $x - 2y = -3$ * $3y = -4x + 10$ and $y = \frac{x}{2} + \frac{3}{2}$ * $y = -\frac{4}{3}x + \frac{10}{3}$ Enter the functions $Y_1 = -4X/3 + 10/3$ and $Y_2 = X/2 + 3/2$....

## Coordinate Geometry (Chapter 6) 141 **Step 1:** Rearrange each equation into the form $y = mx + c$. * $4x + 3y = 10$ and $x - 2y = -3$ * $3y = -4x + 10$ and $y = \frac{x}{2} + \frac{3}{2}$ * $y = -\frac{4}{3}x + \frac{10}{3}$ Enter the functions $Y_1 = -4X/3 + 10/3$ and $Y_2 = X/2 + 3/2$. **Step 2:** Draw the **graphs** of the functions on the same set of axes. You may have to change the viewing window. **Step 3:** Use the built in functions to calculate the point of intersection. In this case, the point of intersection is $(1, 2)$. **What to do:** **1** Use technology to find the point of intersection of: * **a** $y = x + 4$, $5x - 3y = 0$ * **b** $x + 2y = 8$, $y = 7 - 2x$ * **c** $x - y = 5$, $2x + 3y = 4$ * **d** $2x + y = 7$, $3x - 2y = 1$ * **e** $y = 3x - 1$, $3x - y = 6$ * **f** $y = -\frac{2x}{3} + 2$, $2x + 3y = 6$ **2** Comment on the use of technology to find the point(s) of intersection in **1e** and **1f**. ## REVIEW SET 6A **1** For $A(3, -2)$ and $B(1, 6)$ find: * **a** the distance from A to B * **b** the midpoint of [AB] * **c** the gradient of [AB] * **d** the coordinates of C if B is the midpoint of [AC]. **2** $P(3, 1)$ and $Q(-1, a)$ are 5 units apart. Find $a$. **3** Use the distance formula to help classify triangle PQR given the points $P(3, 2)$, $Q(-1, 4)$ and $R(-1, 0)$. **4** Two lines have gradients $-\frac{2}{3}$ and $\frac{a}{6}$. Find $a$ if: * **a** the lines are parallel * **b** the lines are perpendicular. **5** ABCD is a parallelogram. Find: * **a** the coordinates of M where its diagonals meet * **b** the coordinates of A. The parallelogram ABCD is depicted with vertices A, B, C, and D. Point M is located at the intersection of the diagonals. The coordinates of B are (2, 4), C are (0, 1), and D are (-4, 0). --- ## Coordinate Geometry (Chapter 6) 143 **6** The line through (2, 1) and (-3, n) has a gradient of -4. Find n. **7** Find the gradient and y-intercept of the line with equation: * **a** $y = \frac{2}{3}x + 3$ * **b** $2x - 3y = 8$ **8** Find the equations of: * **a** line (1), the x-axis * **b** line (2) * **c** line (3). **9** Find the equation of the line: * **a** with gradient 5 and y-intercept -2 * **b** with gradient $\frac{2}{3}$ passing through (3, -1) * **c** passing through (4, -3) and (-1, 1). **10** Sketch the graph of the line with equation $x - 2y = 6$. **11** Does the point (-3, 4) lie on the line with equation $3x - y = 13$? **12** Find the point of intersection of $2x + y = 10$ and $3x - y = 10$ by graphing the two lines on the same set of axes. **13** The graph alongside indicates the number of litres of water which run from a tank over a period of time. * **a** Find the gradient of the line. * **b** Interpret the gradient found in a. * **c** Is the rate of outflow of water constant or variable? What evidence do you have for your answer? The graph shows outflow (L) on the y-axis ranging from 0 to 4000 and time (h) on the x-axis ranging from 0 to 12. ## REVIEW SET 6B **1** For A(-1, 4) and B(2, -3) find: * **a** the distance from B to A * **b** the midpoint of [AB] * **c** the gradient of [AB] * **d** the equation of the line through A and B. **2** Use distances to classify triangle PQR with P(0, 1), Q(-1, -2) and R(3, -3). **3** A(2, 4) and B(k, -1) are $\sqrt{29}$ units apart. Find k. **4** Solve the **Opening Problem** on page 118. distance travelled (km) (25,350) A (30,300) B fuel consumption (litres) **5** The graphs alongside indicate the distance travelled for different amounts of fuel consumed at speeds of 50 kmh⁻¹ (graph A) and 80 kmh⁻¹ (graph B). **a** Find the gradient of each line. **b** What do these gradients mean? **c** If fuel costs $1.17 per litre, how much more would it cost to travel 1000 km at 80 kmh⁻¹ compared with 50 kmh⁻¹? **6** a Prove that OABC is a rhombus. **b** Use gradients to show that the diagonals [OB] and [AC] are perpendicular. A quadrilateral OABC is displayed, with vertices O(0,0), C(3,4), B(8,4), and A(5,0). **7** Two lines have gradients $\frac{4}{-d}$ and $-\frac{d}{4}$. Find $d$ if the lines are: * **a** perpendicular * **b** parallel. **8** The line through (a, 1) and (2, -1) has a gradient of -2. Find a. **9** Find the gradient and y-intercept of the lines with equations: * **a** $y = \frac{x + 2}{3}$ * **b** $3x + 2y = 8$ **10** Find the equation of the line: * **a** with gradient 5 and y-intercept -1 * **b** with x and y-intercepts 2 and -5 respectively * **c** which passes through (-1, 3) and (2, 1). **11** Find k if (2, k) lies on the line with equation $2x + 7y = 41$. **12** On the same set of axes, graph the lines with equations: * **a** $x = 2$ * **b** $3x + 4y = -12$ **13** **a** On the same set of axes, graph the lines with equations $x + 3y = 9$ and $2x - y = 4$. **b** Find the simultaneous solution of $\begin{cases} x + 3y = 9 \\ 2x - y = 4 \end{cases}$