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Practice Problems Simplified Revision Notes

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Practice Problems

Problems:


Problem 1

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Question: Given the points A(2,3)A(2, 3) and B(5,7) B(5, 7), find the distance between them and the midpoint of the line segment connecting them.

Explanation: Imagine you have two points on a map, and you want to know how far apart they are and where the halfway point is between them. The distance tells you how far apart the points are, and the midpoint tells you where the exact middle is.


Problem 2

infoNote

Question: Given the points C(1,2)C(1, 2) and D(4,8)D(4, 8), find the slope of the line that passes through these points. Then, find the equation of the line.

Explanation: The slope of a line tells us how steep it is, like how a hill slopes up or down. The equation of the line is like a recipe that shows us all the points that the line passes through.


Problem 3

infoNote

Question: Find the equation of a line that has a slope of 33 and crosses the yaxisy-axis at y=2.y = -2.

Explanation: If you know how steep a line is (the slope) and where it crosses the y$$-axis (the yy-intercept), you can easily write the equation of the line. This equation tells you everything you need to know about the line.


Problem 4

infoNote

Question: Find the point of intersection of the lines given by the equations 2x+y=42x + y = 4 and xy=1x - y = 1.

Explanation: When two lines cross each other, the point where they meet is called the intersection. Finding this point helps us understand exactly where the lines overlap on a graph.


Solutions:


Problem 1

infoNote

Question: Given the points A(2,3)A(2, 3) and B(5,7) B(5, 7), find the distance between them and the midpoint of the line segment connecting them.

Solution:

We are given two points, A(2,3)A(2, 3) and B(5,7)B(5, 7), and we need to find two things: the distance between them and the midpoint of the line segment that connects them.

Step 1: Find the Distance

To find the distance between two points on a coordinate plane, we use the distance formula:

Distance AB=(x2x1)2+(y2y1)2\text{Distance } |AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Here's how to use it step by step:

  1. Label the points:
  • Point AA has coordinates x1=2x_1 = 2 and y1=3y_1 = 3.
  • Point BB has coordinates x2=5x_2 = 5 and y2=7y_2 = 7.
  1. Substitute the coordinates into the formula: AB=(52)2+(73)2|AB| = \sqrt{(5 - 2)^2 + (7 - 3)^2}

  2. Simplify the expression:

  • Subtract the xx-coordinates: 52=3 5 - 2 = 3.
  • Subtract the yy-coordinates: 73=47 - 3 = 4. AB=32+42|AB| = \sqrt{3^2 + 4^2}
  1. Square the results:
  • 32=93^2 = 9
  • 42=164^2 = 16 AB=9+16=25|AB| = \sqrt{9 + 16} = \sqrt{25}
  1. Find the square root: AB=5|AB| = 5

So, the distance between points AA and BB is 5 units.

Step 2: Find the Midpoint

To find the midpoint of a line segment, we use the midpoint formula:

Midpoint=(x1+x22,y1+y22)\text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Here's how to use it step by step:

  1. Substitute the coordinates into the formula: Midpoint=(2+52,3+72)\text{Midpoint} = \left(\frac{2 + 5}{2}, \frac{3 + 7}{2}\right)

  2. Add the coordinates and divide by 22: Midpoint=(72,102)=(3.5,5)\text{Midpoint} = \left(\frac{7}{2}, \frac{10}{2}\right) = \left(3.5, 5\right)

So, the midpoint of the line segment connecting AA and BB is (3.5, 5).

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Problem 2

infoNote

Question: Given the points C(1,2)C(1, 2) and D(4,8)D(4, 8), find the slope of the line that passes through these points. Then, find the equation of the line.

Solution:

We are given two points, C(1,2)C(1, 2) and D(4,8)D(4, 8). We need to find the slope of the line passing through these points and then find the equation of that line.

Step 1: Find the Slope

To find the slope of a line between two points, we use the slope formula:

Slope m=y2y1x2x1\text{Slope } m = \frac{y_2 - y_1}{x_2 - x_1}

Here's how to use it step by step:

  1. Label the points:
  • Point CC has coordinates x1=1x_1 = 1 and y1=2y_1 = 2.
  • Point DD has coordinates x2=4x_2 = 4 and y2=8y_2 = 8.
  1. Substitute the coordinates into the formula: m=8241m = \frac{8 - 2}{4 - 1}

  2. Simplify the expression:

  • Subtract the y-coordinates: 82=68 - 2 = 6.
  • Subtract the x-coordinates: 41=34 - 1 = 3. m=63=2m = \frac{6}{3} = 2

So, the slope of the line is 2.

Step 2: Find the Equation of the Line

Now that we have the slope m=2m = 2, we can use the point-slope formula to find the equation of the line. The point-slope formula is:

yy1=m(xx1)y - y_1 = m(x - x_1)

Let's use point C(1,2)C(1, 2):

  1. Substitute the values: y2=2(x1)y - 2 = 2(x - 1)

  2. Expand and simplify:

  • Distribute the 22 on the right-hand side: y2=2x2y - 2 = 2x - 2
  1. Solve for yy:
  • Add 22 to both sides: y=2xy = 2x

So, the equation of the line is y = 2x.

image

Problem 3

infoNote

Question: Find the equation of a line that has a slope of 33 and crosses the yaxisy-axis at y=2y = -2.

Solution:

We are given that the slope of the line is m=3m = 3, and it crosses the yaxisy-axis at y=2y = -2. This means the y-intercept is(0,2) (0, -2).

To find the equation of the line, we use the slope-intercept form of a line:

y=mx+cy = mx + c

Here's how to use it step by step:

  1. Substitute the values:
  • The slope m=3m = 3.
  • The y-intercept c=2.c = -2. y=3x2y = 3x - 2

So, the equation of the line is y = 3x - 2.

image

Problem 4

infoNote

Question: Find the point of intersection of the lines given by the equations 2x+y=42x + y = 4 and xy=1x - y = 1.

Solution:

We are given two equations:

  • 2x+y=42x + y = 4 (EquationAEquation A)
  • xy=1x - y = 1 (EquationBEquation B) We need to find the point where these two lines intersect by solving the equations simultaneously.

Step 1: Add the Equations to Eliminate yy

To eliminate yy, we can add Equation AA and Equation BB. Here's how to do it step by step:

  1. Write down the equations:
  • Equation A: 2x+y=42x + y = 4
  • Equation B: xy=1x - y = 1
  1. Add the equations: (2x+y)+(xy)=4+1(2x + y) + (x - y) = 4 + 1

Simplify:

3x=53x = 5

Step 2: Solve for xx

Now, solve for xx :

x=53x = \frac{5}{3}

Step 3: Substitute x=53x = \frac{5}{3} Back into One of the Original Equations

We'll use Equation BB to find yy:

53y=1\frac{5}{3} - y = 1

Subtract 53\frac{5}{3} from both sides:

y=153=3353=23-y = 1 - \frac{5}{3} = \frac{3}{3} - \frac{5}{3} = \frac{-2}{3}

Finally, multiply both sides by 1-1:

y=23y = \frac{2}{3}

Step 4: Write the Solution as a

Coordinate Pair**

The point of intersection is (53,23)\left(\frac{5}{3}, \frac{2}{3}\right).

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