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Math  /  Algebra

QuestionName:
Ai \qquad \qquad -
Writing Equations of Lines \qquad Exit Ticket 1) 13 marks] AA line passes through the points (3,2)(3,2) and (5,−1)(5,-1).
Find the equation of this line in the form y=mx+by=m x+b. 2) 13 marks/ Find the equation of the line with gradient 23\frac{2}{3} that passes through the point (−2,−1)(-2,-1) in the form ax+by+d=0a x+b y+d=0.

Studdy Solution

STEP 1

1. We are given two problems to solve, each requiring the equation of a line in a specific form.
2. For the first problem, we need to find the equation of a line passing through two given points.
3. For the second problem, we need to find the equation of a line with a given gradient passing through a given point.
4. We will use the point-slope form and slope-intercept form for the first problem, and the standard form for the second problem.

STEP 2

1. Calculate the slope of the line passing through two points.
2. Use the slope and one point to find the equation of the line in slope-intercept form.
3. Use the given gradient and point to find the equation of the line in standard form.

STEP 3

Calculate the slope m m of the line passing through the points (3,2)(3,2) and (5,−1)(5,-1).
The formula for the slope m m is:
m=y2−y1x2−x1 m = \frac{y_2 - y_1}{x_2 - x_1}
Substitute the given points into the formula:
m=−1−25−3=−32 m = \frac{-1 - 2}{5 - 3} = \frac{-3}{2}

STEP 4

Use the slope m=−32 m = -\frac{3}{2} and the point (3,2)(3, 2) to find the equation of the line in the form y=mx+b y = mx + b .
Start with the point-slope form of the equation:
y−y1=m(x−x1) y - y_1 = m(x - x_1)
Substitute the slope and the point (3,2)(3, 2):
y−2=−32(x−3) y - 2 = -\frac{3}{2}(x - 3)

STEP 5

Simplify the equation to get it into the slope-intercept form y=mx+b y = mx + b :
y−2=−32x+92 y - 2 = -\frac{3}{2}x + \frac{9}{2}
Add 2 to both sides:
y=−32x+92+2 y = -\frac{3}{2}x + \frac{9}{2} + 2
Convert 2 to a fraction with a denominator of 2:
y=−32x+92+42 y = -\frac{3}{2}x + \frac{9}{2} + \frac{4}{2}
Combine the fractions:
y=−32x+132 y = -\frac{3}{2}x + \frac{13}{2}

STEP 6

For the second problem, use the given gradient 23\frac{2}{3} and the point (−2,−1)(-2, -1) to find the equation in the form ax+by+d=0 ax + by + d = 0 .
Start with the point-slope form:
y−y1=m(x−x1) y - y_1 = m(x - x_1)
Substitute the gradient and the point:
y+1=23(x+2) y + 1 = \frac{2}{3}(x + 2)

STEP 7

Simplify and rearrange to get the equation in standard form:
y+1=23x+43 y + 1 = \frac{2}{3}x + \frac{4}{3}
Multiply through by 3 to eliminate fractions:
3y+3=2x+4 3y + 3 = 2x + 4
Rearrange to standard form:
2x−3y−1=0 2x - 3y - 1 = 0
The equations of the lines are:
1. y=−32x+132 y = -\frac{3}{2}x + \frac{13}{2}
2. 2x−3y−1=0 2x - 3y - 1 = 0

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