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Geometry and Trigonometry - Straight-line equations

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The gradient (slope) of a straight line, denoted by mm, measures the steepness and direction. It is calculated as the change in yy (vertical rise) divided by the change in xx (horizontal run).

A line segment showing the vertical rise and horizontal run between two points A and B.
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Straight lines can be represented in different forms: Gradient-intercept form y=mx+cy = mx + c, Point-gradient form y−y1=m(x−x1)y - y_1 = m(x - x_1), and General form ax+by+d=0ax + by + d = 0.

Graph of a line highlighting the y-intercept 'c' where the line crosses the y-axis.
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Parallel lines have identical gradients (m1=m2m_1 = m_2), meaning they never intersect. Perpendicular lines intersect at 90∘90^\circ and their gradients are negative reciprocals (m1×m2=−1m_1 \times m_2 = -1).

Two lines L1 and L2 intersecting at the origin at right angles.
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The midpoint MM of a line segment is the average of the coordinates of the endpoints, while the distance dd is the length of the segment calculated using the Pythagorean theorem.

📐Formulae

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

y=mx+cy = mx + c

y−y1=m(x−x1)y - y_1 = m(x - x_1)

ax+by+d=0ax + by + d = 0

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

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

mperp=−1mm_{perp} = -\frac{1}{m}

💡Examples

Problem 1:

Find the equation of the line passing through the points A(2,5)A(2, 5) and B(4,9)B(4, 9). Give your answer in the form y=mx+cy = mx + c.

Solution:

  1. Calculate the gradient: m=9−54−2=42=2m = \frac{9 - 5}{4 - 2} = \frac{4}{2} = 2
  2. Use the point-gradient form with point A(2,5)A(2, 5): y−5=2(x−2)y - 5 = 2(x - 2)
  3. Expand and simplify: y−5=2x−4y - 5 = 2x - 4 y=2x+1y = 2x + 1

Explanation:

First, the gradient mm is found using the formula. Then, we substitute one point and the gradient into the line equation and solve for yy to reach the gradient-intercept form.

Problem 2:

Line L1L_1 has the equation y=3x−4y = 3x - 4. Line L2L_2 is perpendicular to L1L_1 and passes through the point (6,2)(6, 2). Find the equation of L2L_2.

Solution:

  1. Identify the gradient of L1L_1: m1=3m_1 = 3.
  2. Find the perpendicular gradient: m2=−13m_2 = -\frac{1}{3}
  3. Use the point-gradient form: y−2=−13(x−6)y - 2 = -\frac{1}{3}(x - 6)
  4. Distribute the gradient: y−2=−13x+2y - 2 = -\frac{1}{3}x + 2
  5. Add 2 to both sides: y=−13x+4y = -\frac{1}{3}x + 4

Explanation:

Perpendicular lines have gradients that multiply to −1-1. Since L1L_1 has a gradient of 33, L2L_2 must have a gradient of −13-\frac{1}{3}. We then use the given point to determine the specific line equation.

Problem 3:

Calculate the distance between the points P(−1,3)P(-1, 3) and Q(5,11)Q(5, 11).

Solution:

  1. Apply the distance formula: d=(5−(−1))2+(11−3)2d = \sqrt{(5 - (-1))^2 + (11 - 3)^2}
  2. Simplify inside the parentheses: d=(6)2+(8)2d = \sqrt{(6)^2 + (8)^2}
  3. Square the numbers: d=36+64d = \sqrt{36 + 64}
  4. Calculate the sum and square root: d=100=10d = \sqrt{100} = 10

Explanation:

The distance formula measures the length of the interval between two coordinates. Here, the horizontal difference is 66 and the vertical difference is 88, leading to a distance of 1010 units.

Problem 4:

Points A(1,2)A(1, 2) and B(7,10)B(7, 10) form a diameter of a circle. Find the equation of the perpendicular bisector of the segment ABAB.

A line segment AB with its midpoint M and a perpendicular line passing through M.

Solution:

  1. Find the midpoint MM of ABAB: M=(1+72,2+102)=(4,6)M = \left( \frac{1 + 7}{2}, \frac{2 + 10}{2} \right) = (4, 6)
  2. Find the gradient of ABAB: mAB=10−27−1=86=43m_{AB} = \frac{10 - 2}{7 - 1} = \frac{8}{6} = \frac{4}{3}
  3. Find the perpendicular gradient: mperp=−14/3=−34m_{perp} = -\frac{1}{4/3} = -\frac{3}{4}
  4. Use the point-gradient formula with M(4,6)M(4, 6): y−6=−34(x−4)y - 6 = -\frac{3}{4}(x - 4) y−6=−0.75x+3y - 6 = -0.75x + 3 y=−0.75x+9y = -0.75x + 9 or 3x+4y−36=03x + 4y - 36 = 0

Explanation:

A perpendicular bisector must pass through the midpoint of the segment and have a gradient that is the negative reciprocal of the segment's gradient.

Problem 5:

A line L1L_1 passes through the point C(2,1)C(2, 1) and has a gradient of m=12m = \frac{1}{2}. A second line L2L_2 passes through the point D(0,5)D(0, 5) and is perpendicular to L1L_1. Find the point of intersection between L1L_1 and L2L_2.

Coordinate geometry plot showing two perpendicular lines L1 and L2 intersecting at point (2, 1).

Solution:

Step 1: Find the equation of L1L_1 using y−y1=m(x−x1)y - y_1 = m(x - x_1). y−1=12(x−2)y - 1 = \frac{1}{2}(x - 2) y=12x−1+1y = \frac{1}{2}x - 1 + 1 y=0.5xy = 0.5x

Step 2: Find the gradient of L2L_2. Since L2⊥L1L_2 \perp L_1, m2=−1m1m_2 = -\frac{1}{m_1}. m2=−10.5=−2m_2 = -\frac{1}{0.5} = -2

Step 3: Find the equation of L2L_2 using point D(0,5)D(0, 5) and m2=−2m_2 = -2. y=−2x+5y = -2x + 5

Step 4: Solve the simultaneous equations to find the intersection. 0.5x=−2x+50.5x = -2x + 5 2.5x=52.5x = 5 x=2x = 2

Step 5: Substitute x=2x = 2 into L1L_1. y=0.5(2)=1y = 0.5(2) = 1

The intersection point is (2,1)(2, 1).

Explanation:

To find the intersection of two lines, we first determine their individual equations. For L1L_1, we use the point-slope form. For L2L_2, we utilize the property that perpendicular lines have negative reciprocal gradients. Finally, we set the two equations equal to each other to solve for the shared xx and yy coordinates.