### All PSAT Math Resources

## Example Questions

### Example Question #1 : How To Find The Equation Of A Perpendicular Line

What line is perpendicular to through ?

**Possible Answers:**

**Correct answer:**

We need to find the slope of the given equation by converting it to the slope intercept form: .

The slope is and the perpendicular slope would be the opposite reciprocal, or .

The new equation is of the form and we can use the point to calculate . The next step is to convert into the standard form of .

### Example Question #1 : How To Find The Equation Of A Perpendicular Line

What line is perpendicular to and passes through ?

**Possible Answers:**

**Correct answer:**

Perpendicular slopes are opposite reciprocals. The original slope is so the new perdendicular slope is 3.

We plug the point and the slope into the point-slope form of the equation:

to get or in standard form .

### Example Question #11 : Perpendicular Lines

Solve the system of equations for the point of intersection.

**Possible Answers:**

**Correct answer:**

First one needs to use one of the two equations to substitute one of the unknowns.

From the second equation we can derive that *y* = *x* – 3.

Then we substitute what we got into the first equation which gives us: *x* + *x* – 3 = 15.

Next we solve for *x*, so 2*x* = 18 and *x* = 9.

*x* – *y* = 3, so *y *= 6.

These two lines will intersect at the point (9,6).

### Example Question #11 : How To Find The Equation Of A Perpendicular Line

Line A is perpendicular to and passes the point . Find the -intercept of line A.

**Possible Answers:**

**Correct answer:**

We are given an equation of a line and told that line A is perpendicular to it. The slope of the given line is 2. Therefore, the slope of line A must be , since perpendicular lines have slopes that are negative reciprocals of each other.

The equation for line A will therefore take the form , where b is the y-intercept.

Since we are told that it crosses , we can plug in the point and solve for c:

Then the equation becomes .

To find the x-intercept, plug in 0 for y and solve for x:

### Example Question #11 : How To Find The Equation Of A Perpendicular Line

What line is perpendicular to through ?

**Possible Answers:**

**Correct answer:**

The slope of the given line is , and the slope of the perpendicular line is its negative reciprocal, . We take the new slope and the given point and plug them into the slope-intercept form of a line, .

Thus, the perpendicular line has the equation , or in standard form, .

### Example Question #11 : How To Find The Equation Of A Perpendicular Line

In the xy-plane, the equation of the line n is –x+8y=17. If the line m is perpendicular to line n, what is a possible equation of line m?

**Possible Answers:**

y= -1/8x + 5

y= 8x-17

y= -8x + 5

x= -8y + (17/8)

**Correct answer:**

y= -8x + 5

We start by add x to the other side of the equation to get the y by itself, giving us 8y =17 + x. We then divide both sides by 8, giving us y= 17/8 + 1/8x. Since we are looking for the equation of a perpendicular line, we know the slope (the coefficient in front of x) will be the opposite reciprocal of the slope of our line, giving us y= -8x + 5 as the answer.

### Example Question #1 : How To Find The Equation Of A Perpendicular Line

What line is perpendicular to *x* + 3*y* = 6 and travels through point (1,5)?

**Possible Answers:**

*y* = 2*x* + 1

*y* = 6*x* – 3

*y* = 3*x* + 2

*y* = 2/3*x* + 6

*y* = –1/3*x* – 4

**Correct answer:**

*y* = 3*x* + 2

Convert the equation to slope intercept form to get *y* = –1/3*x* + 2. The old slope is –1/3 and the new slope is 3. Perpendicular slopes must be opposite reciprocals of each other: *m*_{1 * }*m*_{2} = –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* = 3*x* + 2

### Example Question #1 : How To Find The Equation Of A Perpendicular Line

What line is perpendicular to and passes through ?

**Possible Answers:**

**Correct answer:**

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 2*x* + *y *= 5 and passes through the point (2,7)?

**Possible Answers:**

–*x*/2 + *y* = 6

*x*/2 – *y* = 6

2*x* + *y* = 7

*x*/2 + *y* = 5

2*x* – *y* = 6

**Correct answer:**

–*x*/2 + *y* = 6

First, put the equation of the line given into slope-intercept form by solving for *y*. You get *y* = -2*x* +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* = 1/2*x* + *b* and solving for *b*, we get *b* = 6. Thus, the equation of the line is *y* = ½*x* + 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?*

**Possible Answers:**

4x **–** 3y = 4

2x + y = 3

2x **–** y = 3

3x + 2y = 4

x **–** y = 3

**Correct answer:**

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.

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