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Geometry and Trigonometry - Gradient

Grade 11IB_AI

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

πŸ”‘Concepts

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The gradient (mm) of a line represents the 'steepness' or rate of change between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). It is calculated as the ratio of the vertical change (rise) to the horizontal change (run).

A line segment showing rise and run between two points A and B on a coordinate plane.
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Parallel lines have identical gradients (m1=m2m_1 = m_2). This means they increase or decrease at the exact same rate and will never intersect.

Two parallel lines with the same slope.
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Perpendicular lines intersect at a 90∘90^\circ angle. Their gradients are negative reciprocals of each other (m1Γ—m2=βˆ’1m_1 \times m_2 = -1).

Two lines intersecting at a right angle.
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The gradient is also the tangent of the angle θ\theta that the line makes with the positive xx-axis: m=tan⁑(θ)m = \tan(\theta).

πŸ“Formulae

m=y2βˆ’y1x2βˆ’x1m = \frac{y_2 - y_1}{x_2 - x_1}

m1=m2Β (ParallelΒ Lines)m_1 = m_2 \text{ (Parallel Lines)}

m1Γ—m2=βˆ’1Β (PerpendicularΒ Lines)m_1 \times m_2 = -1 \text{ (Perpendicular Lines)}

m=tan⁑(θ)m = \tan(\theta)

πŸ’‘Examples

Problem 1:

Find the gradient of the line passing through the points A(2,βˆ’3)A(2, -3) and B(5,6)B(5, 6).

Solution:

Using the gradient formula m=y2βˆ’y1x2βˆ’x1m = \frac{y_2 - y_1}{x_2 - x_1}:

m=6βˆ’(βˆ’3)5βˆ’2m=6+33m=93m=3\begin{array}{r} m = \frac{6 - (-3)}{5 - 2} \\ m = \frac{6 + 3}{3} \\ m = \frac{9}{3} \\ m = 3 \end{array}

Explanation:

Substitute the coordinates (x1,y1)=(2,βˆ’3)(x_1, y_1) = (2, -3) and (x2,y2)=(5,6)(x_2, y_2) = (5, 6) into the formula. Remember that subtracting a negative number becomes addition.

Problem 2:

Given a line L1L_1 with the equation y=βˆ’23x+5y = -\frac{2}{3}x + 5, find the gradient of a line L2L_2 that is perpendicular to L1L_1.

Solution:

The gradient of L1L_1 is m1=βˆ’23m_1 = -\frac{2}{3}.

For perpendicular lines, m1Γ—m2=βˆ’1m_1 \times m_2 = -1:

βˆ’23Γ—m2=βˆ’1m2=βˆ’1βˆ’23m2=32\begin{array}{r} -\frac{2}{3} \times m_2 = -1 \\ m_2 = \frac{-1}{-\frac{2}{3}} \\ m_2 = \frac{3}{2} \end{array}

Explanation:

Identify the gradient of the first line from the y=mx+cy = mx + c form. The perpendicular gradient is the negative reciprocal: flip the fraction and change the sign.

Problem 3:

A line makes an angle of 45∘45^\circ with the positive xx-axis. Determine its gradient.

Solution:

Using the relationship between the angle and the gradient:

m=tan⁑(45∘)m = \tan(45^\circ) m=1m = 1

Explanation:

The gradient is equal to the tangent of the angle of inclination.

Problem 4:

Find the gradient of a line that is perpendicular to the line segment CDCD where CC is (βˆ’2,4)( -2, 4) and DD is (4,1)(4, 1).

Line segment CD connecting (-2, 4) and (4, 1).

Solution:

  1. Calculate the gradient of segment CDCD (mCDm_{CD}): mCD=1βˆ’44βˆ’(βˆ’2)=βˆ’36=βˆ’12m_{CD} = \frac{1 - 4}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}
  2. Use the perpendicular gradient property m1Γ—m2=βˆ’1m_1 \times m_2 = -1: βˆ’12Γ—mβŠ₯=βˆ’1-\frac{1}{2} \times m_{\perp} = -1
  3. Solve for mβŠ₯m_{\perp}: mβŠ₯=2m_{\perp} = 2

Explanation:

To find the perpendicular gradient, calculate the slope of the original segment and then take the negative reciprocal (flipthefractionandchangethesignflip the fraction and change the sign).

Problem 5:

Given two parallel lines, L1L_1 and L2L_2, where L1L_1 passes through the points P(1,1)P(1, 1) and Q(4,7)Q(4, 7). If L2L_2 passes through the point R(0,βˆ’2)R(0, -2), find the gradient of L2L_2 and its equation in the form y=mx+cy = mx + c.

Graph showing two parallel lines L1 and L2. L1 passes through (1,1) and (4,7). L2 passes through (0,-2).

Solution:

  1. Calculate the gradient of line L1L_1 using the gradient formula: m1=y2βˆ’y1x2βˆ’x1=7βˆ’14βˆ’1=63=2m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 1}{4 - 1} = \frac{6}{3} = 2

  2. Since L1L_1 and L2L_2 are parallel, their gradients are equal: m2=m1=2m_2 = m_1 = 2

  3. Use the point-slope form or y=mx+cy = mx + c with point R(0,βˆ’2)R(0, -2) to find the equation of L2L_2: Since RR is on the yy-axis, the yy-intercept cc is βˆ’2-2. y=2xβˆ’2y = 2x - 2

Explanation:

Parallel lines share the same steepness, meaning their gradients are identical. Once the gradient of the first line is found using the coordinates of two points, it can be applied directly to the second line. The equation is then formed using the known point and the shared gradient.