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Coordinate Geometry - Equation of a Straight Line (y = mx + c)

Grade 12A Level

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

🔑Concepts

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The equation of a straight line is represented as y=mx+cy = mx + c, where mm is the gradient (slope) and cc is the yy-intercept (the point where the line crosses the vertical axis).

Graph showing a line with gradient m and y-intercept c
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Parallel lines have identical gradients (m1=m2m_1 = m_2). This means they will never intersect, regardless of their yy-intercepts.

Two parallel lines with the same slope
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Perpendicular lines meet at a 90∘90^{\circ} angle. Their gradients are negative reciprocals of each other, satisfying the condition m1×m2=−1m_1 \times m_2 = -1.

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Horizontal lines have a gradient of 00 and are written as y=ky = k. Vertical lines have an undefined gradient and are written as x=kx = k.

📐Formulae

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

Equation of a straight line (Slope-intercept form): y=mx+cy = mx + c

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

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

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

Perpendicular gradient: m⊥=−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(4,13)B(4, 13).

Solution:

m=13−54−2=82=4m = \frac{13 - 5}{4 - 2} = \frac{8}{2} = 4. Using y=mx+cy = mx + c: 5=4(2)+c⇒5=8+c⇒c=−35 = 4(2) + c \Rightarrow 5 = 8 + c \Rightarrow c = -3. Equation: y=4x−3y = 4x - 3.

Explanation:

First, calculate the gradient using the two-point formula. Then, substitute one point and the gradient into the slope-intercept form to solve for the y-intercept (c).

Problem 2:

Find the equation of the line perpendicular to y=2x+5y = 2x + 5 that passes through the point (6,2)(6, 2).

Solution:

Gradient of given line m1=2m_1 = 2. Perpendicular gradient m2=−12m_2 = -\frac{1}{2}. Using y−y1=m(x−x1)y - y_1 = m(x - x_1): y−2=−12(x−6)⇒y−2=−0.5x+3⇒y=−0.5x+5y - 2 = -\frac{1}{2}(x - 6) \Rightarrow y - 2 = -0.5x + 3 \Rightarrow y = -0.5x + 5.

Explanation:

Perpendicular lines have negative reciprocal gradients. After finding the new gradient, use the point-slope formula with the given coordinate.

Problem 3:

A line segment PQPQ has endpoints P(−2,3)P(-2, 3) and Q(4,7)Q(4, 7). Find the equation of the perpendicular bisector of PQPQ.

Solution:

Midpoint M=(−2+42,3+72)=(1,5)M = (\frac{-2+4}{2}, \frac{3+7}{2}) = (1, 5). Gradient PQ=7−34−(−2)=46=23PQ = \frac{7-3}{4-(-2)} = \frac{4}{6} = \frac{2}{3}. Perpendicular gradient m=−32m = -\frac{3}{2}. Equation: y−5=−32(x−1)⇒2y−10=−3x+3⇒3x+2y=13y - 5 = -\frac{3}{2}(x - 1) \Rightarrow 2y - 10 = -3x + 3 \Rightarrow 3x + 2y = 13.

Explanation:

The perpendicular bisector passes through the midpoint of the segment at a right angle. Find the midpoint, the gradient of the segment, the perpendicular gradient, and then the equation.

Problem 4:

Find the equation of the line that passes through the point (0,−4)(0, -4) and is parallel to the line y=3x+7y = 3x + 7. Give your answer in the form y=mx+cy = mx + c.

Graph showing two parallel lines with gradient 3, one passing through (0, -4).

Solution:

1. Parallel lines have the same gradient. Therefore, m=3.1. \text{ Parallel lines have the same gradient. Therefore, } m = 3. 2. The line passes through (0,−4), which is the y-intercept (c).2. \text{ The line passes through } (0, -4) \text{, which is the } y\text{-intercept } (c). 3. Thus, c=−4.3. \text{ Thus, } c = -4. 4. Substitute m and c into y=mx+c.4. \text{ Substitute } m \text{ and } c \text{ into } y = mx + c. y=3x−4y = 3x - 4

Explanation:

Since the line is parallel to y=3x+7y = 3x + 7, it must share the same gradient (m=3m = 3). The point (0,−4)(0, -4) lies on the yy-axis, identifying −4-4 as the yy-intercept (cc).

Problem 5:

A line LL crosses the xx-axis at (5,0)(5, 0) and the yy-axis at (0,2)(0, 2). Find the equation of line LL.

Graph of the line crossing the x-axis at 5 and the y-axis at 2.

Solution:

1. Identify the points: (x1,y1)=(5,0) and (x2,y2)=(0,2).1. \text{ Identify the points: } (x_1, y_1) = (5, 0) \text{ and } (x_2, y_2) = (0, 2). 2. Calculate the gradient m:2. \text{ Calculate the gradient } m: m=2−00−5=2−5=−0.4m = \frac{2 - 0}{0 - 5} = \frac{2}{-5} = -0.4 3. Identify the y-intercept c from the point (0,2):3. \text{ Identify the } y\text{-intercept } c \text{ from the point } (0, 2): c=2c = 2 4. Write the equation:4. \text{ Write the equation:} y=−0.4x+2y = -0.4x + 2

Explanation:

To find the equation, we first calculate the gradient using the change in yy over the change in xx. Since the line crosses the yy-axis at 2, we can directly identify cc as 2.