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Coordinate Geometry - Parallel and Perpendicular Lines

Grade 9IGCSE

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

🔑Concepts

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Parallel lines have identical gradients. If two lines L1L_1 and L2L_2 are parallel, their slopes satisfy m1=m2m_1 = m_2. Visually, these lines never intersect and remain a constant distance apart.

Graph showing two parallel lines with the same gradient.
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Perpendicular lines meet at a right angle (90∘90^\circ). The product of their gradients is −1-1, expressed as m1×m2=−1m_1 \times m_2 = -1. One gradient is the negative reciprocal of the other: m2=−1m1m_2 = -\frac{1}{m_1}.

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To find the equation of a line passing through (x1,y1)(x_1, y_1) that is parallel or perpendicular to a given line, first identify the target gradient mm, then use the point-gradient formula y−y1=m(x−x1)y - y_1 = m(x - x_1).

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Horizontal lines (gradient 00) have the form y=ky = k. Lines perpendicular to them are vertical lines (gradient undefined) which have the form x=hx = h.

📐Formulae

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

y=mx+cy = mx + c (Gradient-intercept form)

m1=m2m_1 = m_2 (Condition for Parallel Lines)

m1×m2=−1m_1 \times m_2 = -1 or m2=−1m1m_2 = -\frac{1}{m_1} (Condition for Perpendicular Lines)

y−y1=m(x−x1)y - y_1 = m(x - x_1) (Point-gradient form)

💡Examples

Problem 1:

Find the equation of the line parallel to y=3x−5y = 3x - 5 that passes through the point (2,10)(2, 10).

Solution:

y=3x+4y = 3x + 4

Explanation:

  1. Since the lines are parallel, they have the same gradient, so m=3m = 3. 2. Use the point (2,10)(2, 10) in the equation y=mx+cy = mx + c: 10=3(2)+c10 = 3(2) + c. 3. Solve for cc: 10=6+c⇒c=410 = 6 + c \Rightarrow c = 4. 4. Write the final equation: y=3x+4y = 3x + 4.

Problem 2:

Line L1 has the equation y=−2x+7y = -2x + 7. Find the equation of line L2 which is perpendicular to L1 and passes through the point (−4,1)(-4, 1).

Solution:

y=12x+3y = \frac{1}{2}x + 3

Explanation:

  1. The gradient of L1 is m1=−2m_1 = -2. 2. The perpendicular gradient m2m_2 is the negative reciprocal: m2=−1/(−2)=1/2m_2 = -1 / (-2) = 1/2. 3. Use y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (−4,1)(-4, 1): y−1=12(x−(−4))y - 1 = \frac{1}{2}(x - (-4)). 4. Simplify: y−1=12x+2⇒y=12x+3y - 1 = \frac{1}{2}x + 2 \Rightarrow y = \frac{1}{2}x + 3.

Problem 3:

Determine if the lines 2x+y=52x + y = 5 and 4x+2y=104x + 2y = 10 are parallel, perpendicular, or the same line.

Solution:

The lines are the same (coincident).

Explanation:

  1. Rearrange both into y=mx+cy = mx + c form. 2. Line 1: y=−2x+5y = -2x + 5. 3. Line 2: 2y=−4x+10⇒y=−2x+52y = -4x + 10 \Rightarrow y = -2x + 5. 4. Since both the gradient (m=−2m = -2) and the y-intercept (c=5c = 5) are identical, they represent the same line.

Problem 4:

Find the equation of the line L1L_1 that passes through the point P(4,3)P(4, 3) and is perpendicular to the line L2L_2 given by the equation y=2x−1y = 2x - 1. Express your answer in the form y=mx+cy = mx + c.

Graph showing two perpendicular lines intersecting. L1 has a negative slope passing through (4,3) and L2 has a positive slope.

Solution:

m2=2m_2 = 2 m1=−1m2=−12m_1 = -\frac{1}{m_2} = -\frac{1}{2} y−y1=m(x−x1)y - y_1 = m(x - x_1) y−3=−12(x−4)y - 3 = -\frac{1}{2}(x - 4) y−3=−12x+2y - 3 = -\frac{1}{2}x + 2 y=−12x+5y = -\frac{1}{2}x + 5

Explanation:

First, identify the gradient of the given line L2L_2, which is m=2m = 2. For perpendicular lines, the product of their gradients is −1-1, so the gradient of L1L_1 is the negative reciprocal, m=−12m = -\frac{1}{2}. Use the point-slope formula with the coordinates (4,3)(4, 3) and the new gradient to find the equation of the line, then simplify to the gradient-intercept form.

Problem 5:

Line AA passes through the points (0,−2)(0, -2) and (4,0)(4, 0). Line BB is parallel to Line AA and passes through the point (0,3)(0, 3). Find the equation of Line BB.

Graph showing two parallel lines with positive gradients. Line A passes through (0,-2) and (4,0). Line B passes through (0,3).

Solution:

mA=0−(−2)4−0=24=0.5m_A = \frac{0 - (-2)}{4 - 0} = \frac{2}{4} = 0.5 mB=mA=0.5m_B = m_A = 0.5 y=mx+cy = mx + c y=0.5x+3y = 0.5x + 3

Explanation:

To find the gradient of Line AA, use the gradient formula with points (0,−2)(0, -2) and (4,0)(4, 0), which gives m=0.5m = 0.5. Since Line BB is parallel to Line AA, it must have the same gradient. Line BB passes through (0,3)(0, 3), which is the yy-intercept (c=3c = 3). Substitute these values into the slope-intercept form.