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Chapter 1: Problem 9

Plot the following points on graph paper: $$ \mathrm{P}(4,0), \mathrm{Q}(-2,9), \mathrm{R}(5,8), \mathrm{S}(-1,-2) $$ Hence find the coordinates of the point of intersection of the line passing through \(\mathrm{P}\) and \(\mathrm{Q}\), and the line passing through \(\mathrm{R}\) and \(\mathrm{S}\).

Short Answer

Expert verified
The intersection point is \(-2, 9)\.

Step by step solution

01

- Plot Point P(4,0)

Locate the point \(4,0\) on the graph paper. This point is on the x-axis, 4 units to the right of the origin.
02

- Plot Point Q(-2,9)

Locate the point \(-2, 9\) on the graph paper. This point is 2 units to the left of the origin on the x-axis and 9 units up on the y-axis.
03

- Plot Point R(5,8)

Locate the point \(5,8\) on the graph paper. This point is 5 units to the right on the x-axis and 8 units up on the y-axis.
04

- Plot Point S(-1,-2)

Locate the point \(-1,-2\) on the graph paper. This point is 1 unit to the left on the x-axis and 2 units down on the y-axis.
05

- Draw Line PQ

Draw a straight line passing through points \(4,0\) and \(-2,9\).
06

- Draw Line RS

Draw a straight line passing through points \(5,8\) and \(-1,-2\).
07

- Find the Intersection

Calculate the intersection of the lines. Find the equations of the lines to solve for the intersection point. For line \(PQ\): \[ \frac{y-0}{9-0} = \frac{x-4}{-2-4} \] simplifying, we get: \[ y = -\frac{3}{2}x + 6 \] For line \(RS\): \[ \frac{y-8}{-2-8} = \frac{x-5}{-1-5} \] simplifying, we get: \[ y = \frac{10}{6} x - \frac{10}{6}(5) + 8 = \frac{5}{3}x - \frac{34}{3} \] Equate the two line equations to find the intersection point: \[ -\frac{3}{2}x + 6 = \frac{5}{3}x - \frac{34}{3} \] Solve for \(x\), then substitute back to find \(y\).
08

- Solve for x

Equate and solve the x-values: \[ -\frac{3}{2}x + 6 = \frac{5}{3}x - \frac{34}{3} \] Combining the terms: \[ -\frac{3}{2}x - \frac{5}{3}x = -\frac{130}{6} - 12 \] Simplifying, we get \[ x = -2 \].
09

- Solve for y

Substitute \(x = -2\) back into one of the line equations: \[ y = -\frac{3}{2} (-2) + 6 = 3 + 6 = 9 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graph Plotting
Graph plotting involves placing points on a coordinate system. Each point has an x-coordinate (horizontal axis) and a y-coordinate (vertical axis). For instance, to plot point \(\text{P}(4,0)\), move 4 units to the right on the x-axis and stay at 0 on the y-axis. For point \(\text{Q}(-2,9)\), move 2 units to the left on the x-axis and 9 units up on the y-axis. Plotting correctly is crucial for visualizing relationships and solving problems. Remember:
  • Start at the origin (0,0).
  • Move horizontally to the x-coordinate.
  • Move vertically to the y-coordinate.
Line Equations
A line equation represents all the points on a line. It is often written in the form \(y = mx + c\), where \m\ is the slope (steepness) of the line, and \c\ is the y-intercept (where the line crosses the y-axis). To find a line equation passing through two points:
  • Calculate the slope: \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
  • Use one point to solve for the y-intercept \(c\).
  • Combine these to write the equation.
Point of Intersection
The point of intersection is where two lines cross. To find this:
  • First, get the equations of both lines.
  • Set the equations equal to each other and solve for \x\.
  • Substitute this \x\ value back into one of the equations to find the \y\ value.
For example, solving \(-\frac{3}{2}x + 6 = \frac{5}{3}x - \frac{34}{3}\), we find \x = -2\. Substituting \x = -2\ back into one equation gives \y = 9\. Hence, the intersection is at \(-2, 9\).
Coordinate Systems
A coordinate system is a grid formed by two perpendicular lines: the x-axis (horizontal) and the y-axis (vertical). The point where they intersect is the origin \((0,0)\). The coordinate system divides the plane into four quadrants, helping locate points. Each point is represented as \((x,y)\):
  • The first number is the x-coordinate (horizontal position).
  • The second is the y-coordinate (vertical position).
This system is fundamental for graphing functions and solving geometric problems.

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Most popular questions from this chapter

(1) Without using a calculator evaluate (a) \(\frac{1}{2} \times \frac{3}{4}\) (b) \(7 \times \frac{1}{14}\) (c) \(\frac{2}{3} \div \frac{8}{9}\) (d) \(\frac{8}{9} \div 16\) (2) Confirm your answer to part (1) using a calculator.

If \(f(x)=3 x+15\) and \(g(x)=1 / 3 x-5\), evaluate (a) \(f(2)\) (b) \(f(10)\) (c) \(f(0)\) (d) \(g(21)\) (e) \(g(45)\) (f) \(g(15)\) What word describes the relationship between \(f\) and \(g\) ?

(Excel) The supply and demand functions of a good are given by $$ \begin{aligned} &P=-Q_{\mathrm{D}}+240 \\ &P=60+2 Q_{\mathrm{s}} \end{aligned} $$ where \(P, Q_{D}\) and \(Q_{5}\) denote price, quantity demanded and quantity supplied, respectively. Sketch graphs of both functions on the same diagram, on the range \(0 \leq Q \leq 80\) and hence find the equilibrium price. The government now imposes a fixed tax, \(\$ 60\), on each good. Draw the new supply equation on the same diagram and hence find the new equilibrium price. What fraction of the \(\$ 60\) tax is paid by the consumer? Consider replacing the demand function by the more general equation $$ P=-k Q_{\mathrm{D}}+240 $$ By repeating the calculations above, find the fraction of the tax paid by the consumer for the case when \(k\) is (a) 2 (b) 3 (c) 4 State the connection between this fraction and the value of \(k\). Use this connection to predict how much tax is paid by the consumer when \(k=6\).

The demand and supply functions for two interdependent commodities are given by $$ \begin{aligned} &Q_{\mathrm{D}_{1}}=40-5 P_{1}-P_{2} \\ &Q_{\mathrm{D}_{2}}=50-2 P_{1}-4 P_{2} \\ &Q_{\mathrm{s}_{1}}=-3+4 P_{1} \\ &Q_{\mathrm{s}_{2}}=-7+3 P_{2} \end{aligned} $$ where \(Q_{D_{i}}, Q_{s,}\) and \(P_{i}\) denote the quantity demanded, quantity supplied and price of good \(i\) respectively. Determine the equilibrium price and quantity for this two-commodity model. Are these goods substitutable or complementary?

Given that consumption, investment, \(C=0.8 Y+60\) \(I=-30 r+740\) \(M_{\mathrm{S}}=4000\) money supply, transaction-precautionary demand for money, \(M_{\mathrm{S}}=0.15 Y\) speculative demand for money, \(\quad L_{2}=-20 r+3825\) determine the values of national income, \(Y\), and interest rate, \(r\), on the assumption that both the commodity and the money markets are in equilibrium.
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