All SSAT Upper Level Math Resources
Example Questions
Example Question #1 : How To Find The Equation Of A Perpendicular Line
What line is perpendicular to x + 3y = 6 and travels through point (1,5)?
y = 2x + 1
y = 2/3x + 6
y = 3x + 2
y = 6x – 3
y = –1/3x – 4
y = 3x + 2
Convert the equation to slope intercept form to get y = –1/3x + 2. The old slope is –1/3 and the new slope is 3. Perpendicular slopes must be opposite reciprocals of each other: m1 * m2 = –1
With the new slope, use the slope intercept form and the point to calculate the intercept: y = mx + b or 5 = 3(1) + b, so b = 2
So y = 3x + 2
Example Question #1 : How To Find The Equation Of A Perpendicular Line
What line is perpendicular to and passes through
?
Convert the given equation to slope-intercept form.
The slope of this line is . The slope of the line perpendicular to this one will have a slope equal to the negative reciprocal.
The perpendicular slope is .
Plug the new slope and the given point into the slope-intercept form to find the y-intercept.
So the equation of the perpendicular line is .
Example Question #1 : How To Find The Equation Of A Perpendicular Line
What is the equation of a line that runs perpendicular to the line 2x + y = 5 and passes through the point (2,7)?
2x – y = 6
–x/2 + y = 6
2x + y = 7
x/2 + y = 5
x/2 – y = 6
–x/2 + y = 6
First, put the equation of the line given into slope-intercept form by solving for y. You get y = -2x +5, so the slope is –2. Perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is 1/2. Plugging in the point given into the equation y = 1/2x + b and solving for b, we get b = 6. Thus, the equation of the line is y = ½x + 6. Rearranged, it is –x/2 + y = 6.
Example Question #1 : How To Find The Equation Of A Perpendicular Line
Line m passes through the points (1, 4) and (5, 2). If line p is perpendicular to m, then which of the following could represent the equation for p?
4x – 3y = 4
3x + 2y = 4
2x + y = 3
2x – y = 3
x – y = 3
2x – y = 3
The slope of m is equal to y2-y1/x2-x1 = 2-4/5-1 = -1/2
Since line p is perpendicular to line m, this means that the products of the slopes of p and m must be –1:
(slope of p) * (-1/2) = -1
Slope of p = 2
So we must choose the equation that has a slope of 2. If we rewrite the equations in point-slope form (y = mx + b), we see that the equation 2x – y = 3 could be written as y = 2x – 3. This means that the slope of the line 2x – y =3 would be 2, so it could be the equation of line p. The answer is 2x – y = 3.
Example Question #1 : How To Find The Equation Of A Perpendicular Line
What is the equation for the line that is perpendicular to through point
?
Perpendicular slopes are opposite reciprocals.
The given slope is found by converting the equation to the slope-intercept form.
The slope of the given line is and the perpendicular slope is
.
We can use the given point and the new slope to find the perpendicular equation. Plug in the slope and the given coordinates to solve for the y-intercept.
Using this y-intercept in slope-intercept form, we get out final equation: .
Example Question #1 : How To Find The Equation Of A Perpendicular Line
Which line below is perpendicular to ?
The definition of a perpendicular line is one that has a negative, reciprocal slope to another.
For this particular problem, we must first manipulate our initial equation into a more easily recognizable and useful form: slope-intercept form or .
According to our formula, our slope for the original line is
. We are looking for an answer that has a perpendicular slope, or an opposite reciprocal. The opposite reciprocal of
is
. Flip the original and multiply it by
.
Our answer will have a slope of . Search the answer choices for
in the
position of the
equation.
is our answer.
(As an aside, the negative reciprocal of 4 is . Place the whole number over one and then flip/negate. This does not apply to the above problem, but should be understood to tackle certain permutations of this problem type where the original slope is an integer.)
Example Question #7 : How To Find The Equation Of A Perpendicular Line
If a line has an equation of , what is the slope of a line that is perpendicular to the line?
Putting the first equation in slope-intercept form yields .
A perpendicular line has a slope that is the negative inverse. In this case, .
Example Question #1 : How To Find The Equation Of A Perpendicular Line
Given a line defined by the equation
, which of the following lines is perpendicular to
?
For a given line defined by the equation
, any line perpendicular to
must have a slope that is the negative reciprocal of
's slope
,
.
In this instance, the slope of line is
, so
. The only line provided with an equation that has this slope is
.
Example Question #2 : How To Find The Equation Of A Perpendicular Line
A given line is defined by the equation
. What is the slope of any line that is perpendicular to
?
Not enough information provided
For a given line defined by the equation
, any line perpendicular to
must have a slope that is the negative reciprocal of
's slope
,
.
Since in this case ,
.
Example Question #11 : How To Find The Equation Of A Perpendicular Line
Which of the following equations represents a line that goes through the point and is perpendicular to the line
?
In order to solve this problem, we need first to transform the equation from standard form to slope-intercept form:
Transform the original equation to find its slope.
First, subtract from both sides of the equation.
Simplify and rearrange.
Next, divide both sides of the equation by 6.
The slope of our first line is equal to . Perpendicular lines have slopes that are opposite reciprocals of each other; therefore, if the slope of one is x, then the slope of the other is equal to the following:
Let's calculate the opposite reciprocal of our slope:
The slope of our line is equal to 2. We now have the following partial equation:
We are missing the y-intercept, . Substitute the x- and y-values in the given point
to solve for the missing y-intercept.
Add 4 to both sides of the equation.
Substitute this value into our partial equation to construct the equation of our line:
All SSAT Upper Level Math Resources
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