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Coordinate Geometry - Slope-Intercept form: y = mx + c-advanced

Grade 9CBSE

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

🔑Concepts

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The slope-intercept form of a linear equation is written as y=mx+cy = mx + c, where mm represents the slope (steepness) and cc represents the yy-intercept (the point where the line crosses the yy-axis).

Graph of y = 0.5x + 2 showing slope and y-intercept.
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The slope mm is defined as the 'rise over run', calculated as y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}. It represents the rate of change of yy with respect to xx.

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If two lines are parallel, their slopes are equal (m1=m2m_1 = m_2). If two lines are perpendicular, the product of their slopes is −1-1 (m1×m2=−1m_1 \times m_2 = -1).

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The xx-intercept can be found by setting y=0y = 0 in the equation and solving for xx, giving the point where the line crosses the horizontal axis.

📐Formulae

y=mx+cy = mx + c

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

m=tan⁡(θ) (where θ is the angle with the positive x-axis)m = \tan(\theta) \text{ (where } \theta \text{ is the angle with the positive x-axis)}

mparallel=mm_{parallel} = m

mperpendicular=−1mm_{perpendicular} = -\frac{1}{m}

💡Examples

Problem 1:

Convert the equation 5x−2y+10=05x - 2y + 10 = 0 into slope-intercept form and find the slope and yy-intercept.

Solution:

Given: 5x−2y+10=05x - 2y + 10 = 0 Step 1: Isolate the yy term: −2y=−5x−10-2y = -5x - 10 Step 2: Divide every term by −2-2: y=−5−2x−10−2y = \frac{-5}{-2}x - \frac{10}{-2} y=52x+5y = \frac{5}{2}x + 5 Comparing with y=mx+cy = mx + c, we find: Slope m=52m = \frac{5}{2} yy-intercept c=5c = 5

Explanation:

To transform the general form to slope-intercept form, we solve for yy so that the coefficient of yy is +1+1.

Problem 2:

Find the equation of a line that passes through the point (4,−1)(4, -1) and is parallel to the line y=3x+7y = 3x + 7.

Solution:

  1. The given line is y=3x+7y = 3x + 7. Its slope m1=3m_1 = 3.
  2. Since the required line is parallel, its slope m2m_2 must also be 33.
  3. Use the point-slope form with m=3m = 3 and point (x1,y1)=(4,−1)(x_1, y_1) = (4, -1): y−y1=m(x−x1)y - y_1 = m(x - x_1) y−(−1)=3(x−4)y - (-1) = 3(x - 4) y+1=3x−12y + 1 = 3x - 12 y=3x−13y = 3x - 13

Explanation:

Parallel lines share the same slope. Once the slope is identified, we use the coordinates of the given point to find the new yy-intercept.

Problem 3:

Determine if the line passing through (0,2)(0, 2) and (2,3)(2, 3) is perpendicular to the line y=−2x+4y = -2x + 4.

Solution:

Step 1: Find the slope m1m_1 of the line passing through (0,2)(0, 2) and (2,3)(2, 3): m1=3−22−0=12m_1 = \frac{3 - 2}{2 - 0} = \frac{1}{2} Step 2: Identify the slope m2m_2 of the line y=−2x+4y = -2x + 4: m2=−2m_2 = -2 Step 3: Check the product of the slopes: m1×m2=12×(−2)=−1m_1 \times m_2 = \frac{1}{2} \times (-2) = -1 Since the product is −1-1, the lines are perpendicular.

Explanation:

The condition for perpendicularity is that the product of the slopes of the two lines must be −1-1.

Problem 4:

Find the equation of a line that passes through the point (2,3)(2, 3) and is perpendicular to the line x−2y+4=0x - 2y + 4 = 0. Express the answer in the slope-intercept form y=mx+cy = mx + c.

Graph showing two perpendicular lines intersecting at a point, with the required line having a negative slope.

Solution:

  1. First, convert the given line x−2y+4=0x - 2y + 4 = 0 into slope-intercept form: −2y=−x−4-2y = -x - 4 y=12x+2y = \frac{1}{2}x + 2 The slope of this line (m1m_1) is 12\frac{1}{2}.

  2. For a perpendicular line, the product of slopes is −1-1. Thus, m2=−1m1m_2 = -\frac{1}{m_1}: m2=−11/2=−2m_2 = -\frac{1}{1/2} = -2.

  3. Use the point-slope form with m=−2m = -2 and point (2,3)(2, 3): y−3=−2(x−2)y - 3 = -2(x - 2) y−3=−2x+4y - 3 = -2x + 4 y=−2x+7y = -2x + 7.

Explanation:

To find a perpendicular line, we determine the negative reciprocal of the original slope. We then apply the point-slope formula and rearrange it into the required y=mx+cy = mx + c format.

Problem 5:

Determine the equation of the line that makes an angle of 135∘135^{\circ} with the positive direction of the xx-axis and has a yy-intercept of −3-3.

Graph of the line y = -x - 3 showing a 135 degree angle with the positive x-axis and a y-intercept at -3.

Solution:

  1. The slope mm is given by m=tan⁡(θ)m = \tan(\theta). m=tan⁡(135∘)=−1m = \tan(135^{\circ}) = -1.

  2. The yy-intercept (cc) is given as −3-3.

  3. Substituting m=−1m = -1 and c=−3c = -3 into the slope-intercept form y=mx+cy = mx + c: y=(−1)x+(−3)y = (-1)x + (-3) y=−x−3y = -x - 3.

Explanation:

The angle of inclination directly provides the slope using the tangent function. Combined with the yy-intercept, the equation is formed by direct substitution into the standard slope-intercept formula.