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Finding the Equation of a Line: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Line movement

A line passes through $(-5,3)$ . Moving 4 units right along it raises the height by 6 units. Write its equation in point-slope form using the given point.
1. Hint 1
  The movement supplies the slope needed with the point.
2. Hint 2
  Divide the rise by the run, then insert the point into $y-y_1=m(x-x_1)$ .
Answer

$y-3=\frac32(x+5)$ .

Full solution

The move is a run of 4 and a rise of 6, so the slope is
$m=\frac64$ $m=\frac32$
Using the point $(-5,3)$ gives
$y-3=\frac32(x+5)$
At the given point both sides are zero.
The stated move reaches $(-1,9)$ , where both sides are $6$ , checking the direction and size of the step.

Answer

$y-3=\frac32(x+5)$ .

Key idea

A movement along a line supplies the slope for its point-slope equation.
Problem 1: Mark complete
Problem 2 A new anchor

The line $y-7=\frac12(x+4)$ is to be written in point-slope form using its point whose first coordinate is $2$ . Write that form.
1. Hint 1
  The line stays the same when a different point on it is used.
2. Hint 2
  Find the $y$ -value at $x=2$ , then keep the original slope with that new point.
Answer

$y-10=\frac12(x-2)$ .

Full solution

At $x=2$ ,
$y-7=\frac12(2+4)$ $y=10$
The new point is $(2,10)$ and the slope remains $1/2$ , giving
$y-10=\frac12(x-2)$
Both the original and new forms simplify to $y=x/2+9$ , so they describe the same line.

Answer

$y-10=\frac12(x-2)$ .

Key idea

A point-slope equation rewritten from another point on the same line, with the same slope, describes the same line.
Problem 2: Mark complete
Problem 3 Equal ratios

Write $\frac{y+1}{6}=\frac{x-2}{4}$ in standard form $Ax+By=C$ with integer coefficients, $A$ zero or positive, and no common factor shared by all three coefficients.
1. Hint 1
  Clear the nonzero numerical denominators before collecting terms.
2. Hint 2
  Multiply both sides by 12, expand, and reduce any common factor.
Answer

$3x-2y=8$ .

Full solution

Multiply by 12 and distribute:
$2(y+1)=3(x-2)$ $2y+2=3x-6$
Moving terms gives
$3x-2y=8$
The three coefficients share no factor greater than 1, and the leading coefficient is positive.
The point $(2,-1)$ makes both original ratios zero and satisfies the result.

Answer

$3x-2y=8$ .

Key idea

Clearing denominators and collecting terms gives an equivalent equation in standard form.
Problem 3: Mark complete
Problem 4 Workshop charge

A workshop charges a fixed entry fee plus a fixed charge for each hour. A 2-hour visit costs 13 dollars, and a 5-hour visit costs 22 dollars. Write a point-slope equation for total cost $c$ in dollars against hours $h$ , then find the cost of an 8-hour visit.
1. Hint 1
  The difference in cost over the difference in hours gives the hourly rate.
2. Hint 2
  Use either recorded visit as the fixed point in the equation.
Answer

$c-13=3(h-2)$ , or $c-22=3(h-5)$ ; an 8-hour visit costs 31 dollars.

Full solution

The hourly rate is
$m=\frac{22-13}{5-2}$ $m=3$
Using the 2-hour visit gives
$c-13=3(h-2)$
For 8 hours,
$c-13=3(8-2)$ $c=31$
The equation gives an entry charge of 7 dollars; $7+3(5)=22$ checks the other recorded visit.

Answer

$c-13=3(h-2)$ , or $c-22=3(h-5)$ ; an 8-hour visit costs 31 dollars.

Key idea

Two recorded totals determine a constant rate and a point-slope model.
Problem 4: Mark complete
Problem 5 Coordinate blanks

In record A, $(k-2,5)$ and $(-7,-3)$ must lie on one vertical line. In record B, $(2,j)$ and $(-7,-3)$ must lie on one horizontal line. Find $k$ and $j$ and give the equation for each record.
1. Hint 1
  A vertical line fixes the first coordinate, while a horizontal line fixes the second.
2. Hint 2
  Match the corresponding coordinate in each pair before writing its equation.
Answer

$k=-5$ and record A is $x=-7$ ; $j=-3$ and record B is $y=-3$ .

Full solution

For A, equal first coordinates require
$k-2=-7$ $k=-5$
Its two distinct points are $(-7,5)$ and $(-7,-3)$ , giving $x=-7$ .

For B, equal second coordinates require $j=-3$ , giving $y=-3$ .
The first coordinates $2$ and $-7$ differ, so the points are distinct and determine a horizontal line.

Answer

$k=-5$ and record A is $x=-7$ ; $j=-3$ and record B is $y=-3$ .

Key idea

A coordinate that stays fixed directly supplies a vertical or horizontal line equation.
Problem 5: Mark complete
Problem 6 Survey marks

The figure shows two survey marks. Write the equation of their line in point-slope form using $A$ , then in standard form with reduced integer coefficients and a positive coefficient of $x$ . Check whether $C(1,1)$ lies on it.

Survey marks $A$ and $B$ on the coordinate plane.

Text description of this figure

A coordinate grid with equal unit lengths on both axes and arrowheads at both ends of each axis. The horizontal x-axis runs from negative four to six and the vertical y-axis from negative three to five, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Two points are plotted and labeled with their letters only: point A, three units left of the y-axis and four units above the x-axis, at negative three, four; and point B, five units right of the y-axis and one unit below the x-axis, at five, negative one. No line is drawn, and no coordinate pairs or other points are shown.
1. Hint 1
  Read the two marks before deciding whether the line is vertical.
2. Hint 2
  Find the slope, write the equation using $A$ , then clear any fraction.
Answer

$y-4=-\frac58(x+3)$ ; $5x+8y=17$ ; $C(1,1)$ does not lie on the line.

Full solution

The marks are $A(-3,4)$ and $B(5,-1)$ , with different first coordinates.
Their slope is
$m=\frac{-1-4}{5-(-3)}$ $m=-\frac58$
Hence
$y-4=-\frac58(x+3)$

Multiplying by 8 gives $8y-32=-5x-15$ , so
$5x+8y=17$
Substituting $A$ gives $5(-3)+8(4)=17$ , and $B$ gives
$5(5)+8(-1)=17.$
At $C(1,1)$ , $5(1)+8(1)=13$ , which is not $17$ , so $C$ does not lie on the line.

Answer

$y-4=-\frac58(x+3)$ ; $5x+8y=17$ ; $C(1,1)$ does not lie on the line.

Key idea

A graph can supply the two points for an equation, and substitution checks a further point.
Problem 6: Mark complete
Problem 7 Raised tracing

Every point of the line $y-2=-2(x+1)$ is moved vertically up 3 units, with its first coordinate unchanged. Write the equation of the new line in point-slope form and in standard form with reduced integer coefficients and a positive coefficient of $x$ .
1. Hint 1
  Think about what a vertical move does to the rise and the run between any two points.
2. Hint 2
  Keep the slope, and move the anchor point $(-1,2)$ up 3 units.
Answer

$y-5=-2(x+1)$ , or any point-slope form with slope $-2$ anchored at a point of the new line, such as $y-3=-2(x-0)$ ; $2x+y=3$ .

Full solution

The original equation is anchored at $(-1,2)$ , which moves to $(-1,5)$ .
For any two points, adding 3 to both heights leaves their difference unchanged, and their horizontal difference also stays unchanged.
The new slope is still $-2$ .

Using the moved point gives
$y-5=-2(x+1)$ $2x+y=3$
As a check, the old point $(0,0)$ moves to $(0,3)$ , which satisfies the new equation.

Answer

$y-5=-2(x+1)$ , or any point-slope form with slope $-2$ anchored at a point of the new line, such as $y-3=-2(x-0)$ ; $2x+y=3$ .

Key idea

Moving every point vertically by the same amount preserves slope and changes the anchor height.
Problem 7: Mark complete
Problem 8 A one-point check

A student is asked for the line through $(1,-4)$ with slope $\frac52$ and writes $y+4=\frac25(x-1)$ . Substituting $(1,-4)$ makes both sides zero, so the student concludes the equation is correct. Is the student right? Explain what that substitution shows about the equation, and give the correct equation if the student's is wrong.
1. Hint 1
  Ask whether the given point would still satisfy the equation if a different slope had been written in front of $(x-1)$ .
2. Hint 2
  From $(1,-4)$ , a slope of $\frac52$ means a run of 2 and a rise of 5; find the point that step reaches.
3. Hint 3
  Substitute that point into the student's equation and compare the two sides.
Answer

No; the given point satisfies any point-slope equation built from it. The required line's point $(3,1)$ , or any other except $(1,-4)$ , fails the student's equation ( $5\ne\frac45$ ). The correct equation is $y+4=\frac52(x-1)$ .

Full solution

At $(1,-4)$ the left side is $-4+4=0$ , and the right side is
$\frac25(1-1)=0.$
The right side is zero there for any slope written in front of $(x-1)$ , so this substitution checks the point but not the slope.

From $(1,-4)$ , a slope of $\frac52$ means a run of 2 and a rise of 5, which reaches $(3,1)$ on the required line.
In the student's equation the left side is
$1+4=5$
The right side is
$\frac25(3-1)=\frac45$
The two sides differ, so the student's equation is wrong: its slope has the rise and the run swapped.

With the correct slope the equation is
$y+4=\frac52(x-1)$
At $(3,1)$ both sides are $5$ , and at $(1,-4)$ both sides are $0$ , so it passes both checks.

Answer

No; the given point satisfies any point-slope equation built from it. The required line's point $(3,1)$ , or any other except $(1,-4)$ , fails the student's equation ( $5\ne\frac45$ ). The correct equation is $y+4=\frac52(x-1)$ .

Key idea

The given point satisfies any point-slope equation built from it, so checking the slope needs a second point found from the slope.
Problem 8: Mark complete
Problem 9 Zero coefficients

A student allows both $A$ and $B$ to be zero in $Ax+By=C$ . For each of $0x+0y=6$ and $0x+0y=0$ , describe the set of points that satisfy it, and decide whether it is a line.
1. Hint 1
  Consider what remains after the zero coefficients remove both coordinate terms.
2. Hint 2
  Substitute any point you like into each equation, and notice whether the choice of point changes the result.
Answer

$0x+0y=6$ has no points; $0x+0y=0$ has every point in the plane; neither is a line.

Full solution

The first equation reduces to
$0=6$
This is false for every choice of coordinates, so it has no points.

The second reduces to
$0=0$
This is true for every point in the plane, including points outside any chosen line.
A standard-form equation for a line therefore needs at least one nonzero variable coefficient.

Answer

$0x+0y=6$ has no points; $0x+0y=0$ has every point in the plane; neither is a line.

Key idea

At least one variable coefficient must be nonzero for a standard-form equation to describe one line.
Problem 9: Mark complete
Problem 10 Two heights

Can any number $m$ make $y+1=m(x+2)$ contain both $(-2,-1)$ and $(-2,4)$ ? Explain and give an equation that does contain both points.
1. Hint 1
  The first coordinate is the same in both points.
2. Hint 2
  Substitute the second point into the proposed form and see whether changing $m$ can help.
Answer

No; the line through both points is $x=-2$ .

Full solution

The point $(-2,-1)$ makes both sides zero for every $m$ .
At $(-2,4)$ , however, the equation becomes
$5=m(0)$
Its right side is zero for every $m$ , so it cannot hold.

The points share first coordinate $-2$ and have different heights, so their line is $x=-2$ .
Both points satisfy that equation; a vertical line has undefined slope, so point-slope form does not apply.

Answer

No; the line through both points is $x=-2$ .

Key idea

A vertical line is written by fixing its first coordinate, because its slope is undefined and point-slope form does not apply.
Problem 10: Mark complete

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