krit.club logo

Coordinate Geometry - Gradient and Midpoint of a Line

Grade 12A Level

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

🔑Concepts

•

The gradient (or slope) mm measures the steepness of a line segment. For any two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), it is the ratio of the 'rise' (vertical change) over the 'run' (horizontal change). A positive gradient slopes upwards from left to right, while a negative gradient slopes downwards.

A line segment showing the rise and run between two points to illustrate gradient.
•

The midpoint MM is the exact center point of a line segment connecting (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). It is calculated by taking the average of the xx-coordinates and the average of the yy-coordinates, effectively finding the 'middle' of the horizontal and vertical spans.

A line segment AB with a point M located exactly in the center.
•

Two lines are parallel if they have the same gradient (m1=m2m_1 = m_2). They will never intersect, regardless of how far they are extended.

Two parallel lines with the same slope.
•

Two lines are perpendicular if they intersect at a right angle (90∘90^{\circ}). The product of their gradients is −1-1 (i.e., m1×m2=−1m_1 \times m_2 = -1).

Two lines intersecting at a 90 degree angle.

📐Formulae

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

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

Perpendicular Gradient: m⊥=−1mm_{\perp} = -\frac{1}{m}

💡Examples

Problem 1:

Find the gradient and the midpoint of the line segment connecting the points A(−4,7)A(-4, 7) and B(2,1)B(2, 1).

Solution:

Gradient m=1−72−(−4)=−66=−1m = \frac{1 - 7}{2 - (-4)} = \frac{-6}{6} = -1. Midpoint M=(−4+22,7+12)=(−22,82)=(−1,4)M = \left( \frac{-4 + 2}{2}, \frac{7 + 1}{2} \right) = \left( \frac{-2}{2}, \frac{8}{2} \right) = (-1, 4).

Explanation:

Apply the gradient formula by subtracting the y-coordinates and x-coordinates. For the midpoint, calculate the average of the x-values and the average of the y-values.

Problem 2:

The midpoint of a line XYXY is M(3,−2)M(3, -2). If the coordinates of XX are (7,4)(7, 4), find the coordinates of point YY.

Solution:

3=7+xY2  ⟹  6=7+xY  ⟹  xY=−13 = \frac{7 + x_Y}{2} \implies 6 = 7 + x_Y \implies x_Y = -1. −2=4+yY2  ⟹  −4=4+yY  ⟹  yY=−8-2 = \frac{4 + y_Y}{2} \implies -4 = 4 + y_Y \implies y_Y = -8. Point Y=(−1,−8)Y = (-1, -8).

Explanation:

Use the midpoint formula as an equation where the midpoint is known. Solve for the unknown coordinates xYx_Y and yYy_Y individually.

Problem 3:

Given line L1L_1 passes through (1,2)(1, 2) and (4,8)(4, 8). Find the gradient of a line L2L_2 that is perpendicular to L1L_1.

Solution:

Gradient of L1(m1)=8−24−1=63=2L_1 (m_1) = \frac{8 - 2}{4 - 1} = \frac{6}{3} = 2. Since L2⊥L1L_2 \perp L_1, m2=−1m1=−12m_2 = -\frac{1}{m_1} = -\frac{1}{2}.

Explanation:

First, calculate the gradient of the first line using the two given points. Then, apply the perpendicular gradient rule: take the negative reciprocal of m1m_1.

Problem 4:

Find the gradient of the line segment joining the points P(2,5)P(2, 5) and Q(6,−3)Q(6, -3). Use this to determine if the line is perpendicular to a line with gradient 0.250.25.

Graph of line segment PQ from (2,5) to (6,-3).

Solution:

  1. Identify coordinates: (x1,y1)=(2,5)(x_1, y_1) = (2, 5) and (x2,y2)=(6,−3)(x_2, y_2) = (6, -3).
  2. Calculate gradient mm: m=−3−56−2=−84=−2m = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2
  3. Check perpendicularity: For a line with gradient m2=0.25m_2 = 0.25, the product is m×m2=−2×0.25=−0.5m \times m_2 = -2 \times 0.25 = -0.5.
  4. Since −0.5≠−1-0.5 \neq -1, the lines are not perpendicular.

Explanation:

We first calculate the gradient using the formula. To check if it is perpendicular to another line, we multiply the two gradients; if the result is not −1-1, they are not perpendicular.

Problem 5:

Point CC has coordinates (1,2)(1, 2) and point DD has coordinates (5,10)(5, 10). Find the coordinates of the midpoint MM and show its position on a coordinate plane.

Line segment CD with its midpoint M at (3,6).

Solution:

  1. Identify coordinates: (x1,y1)=(1,2)(x_1, y_1) = (1, 2) and (x2,y2)=(5,10)(x_2, y_2) = (5, 10).
  2. Apply midpoint formula for xx: xM=1+52=62=3x_M = \frac{1 + 5}{2} = \frac{6}{2} = 3
  3. Apply midpoint formula for yy: yM=2+102=122=6y_M = \frac{2 + 10}{2} = \frac{12}{2} = 6
  4. The midpoint MM is (3,6)(3, 6).

Explanation:

The midpoint is found by averaging the horizontal positions and the vertical positions of the endpoints.