SAT Math - How to find the length of a line with distance formula

One line has four collinear points in order from left to right A, B, C, D. If AB = 10’, CD was twice as long as AB, and AC = 25’, how long is AD?

45'

40'

50'

35'

30'

Explanation

AB = 10 ’

BC = AC – AB = 25’ – 10’ = 15’

CD = 2 * AB = 2 * 10’ = 20 ’

AD = AB + BC + CD = 10’ + 15’ + 20’ = 45’

What is the distance between (1, 4) and (5, 1)?

Explanation

Let P1 = (1, 4) and P2 = (5, 1)

Substitute these values into the distance formula:

Actmath_29_372_q6_1_copy

The distance formula is an application of the Pythagorean Theorem: a2 + b2 = c2

What is the distance of the line drawn between points (–1,–2) and (–9,4)?

√5

Explanation

The answer is 10. Use the distance formula between 2 points, or draw a right triangle with legs length 6 and 8 and use the Pythagorean Theorem.

What is the distance between the points and ?

$3\sqrt{5}$

$4\sqrt{2}$

$2\sqrt{29}$

Explanation

Plug the points into the distance formula and simplify:

distance2 = (_x_2 – _x_1)2 + (_y_2 – _y_1)2 = (7 – 3)2 + (2 – 12)2 = 42 + 102 = 116

distance = √116 = √(4 * 29) = 2√29

Steven draws a line that is 13 units long. If $(-4,1)$ is one endpoint of the line, which of the following might be the other endpoint?

$(1,13)$

$(9,14)$

$(5,12)$

$(3,7)$

$(13,13)$

Explanation

The distance formula is $\sqrt{((x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2})}$ .

Plug in $(-4,1)$ with each of the answer choices and solve.

Plug in $(1,13)$ :

$d=\sqrt{(1-(-4))^2+ (13-1)^2} = \sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$

This is therefore the correct answer choice.

What is the distance between $(1,3)\ and\ (5,6)$ ?

$5$

$6$

$4$

$7$

$8$

Explanation

Let $P_{1}(1,3)$ and $P_{2}(5,6)$ and use the distance formula:

$d = \sqrt{(x_{2} - x_{1})^2+(y_{2} - y_{1})^2}$

What is the distance between the point and the origin?

Explanation

The distance between 2 points is found using the distance or Pythagorean Theorem. Because values are squared in the formula, distance can never be a negative value.

Bill gets in his car and drives north for 30 miles at 40 mph. He then turns west and drives 40 mph for 40 minutes. Finally, he goes directly northeast 40 miles in 25 minutes.

Using the total distance traveled as a straight line ("as the crow flies") and the time spent traveling, which of the following is closest to Bill's average speed?

$32\ mph$

$30\ mph$

$35\ mph$

$40\ mph$

$27\ mph$

Explanation

Each part of the problem gives you 2 out of the 3 pieces of the rate/time/distance relationship, thus allowing you to find the third (if needed) by using the equation:

$Distance = rate\cdot time$

The problem is otherwise an application of geometry and the distance formula. We need to find the distance between 2 points, but we need to go step-by-step to find out where the final point is. The first two steps are relatively easy to follow. He travels 30 miles north. Now he turns west and travels for 40 minutes at 40 mph. This is $\frac{2}{3}$ of an hour, so we have $40\cdot \frac{2}{3}=\frac{80}{3}\ miles$ . Thus if we are looking at standard Cartesian coordinates (starting at the origin), we are now at the point $\left ( \frac{80}{3},30 \right )$ .

We are now on the last step: 40 miles northeast. We need to decipher this into - and - coordinate changes. To do this, we think of a triangle. Because we are moving directly northeast, this is a 45 degree angle to the horizontal. We can thus imagine a 45-45-90 triangle with a hypotenuse of 40. Now using the relationships on triangles we have:

$40^{2} = x^{2} + x^{2} = 2x^{2}\Rightarrow 40 = \sqrt{2x^{2}} = x\sqrt{2} \Rightarrow x = \frac{40}{\sqrt{2}} = 20\sqrt{2}$

So the final step moves us $20\sqrt{2}$ up and to the right. Moving this way from the point $\left ( \frac{80}{3},30 \right )$ leaves us at:

$\left ( \frac{80}{3}-20\sqrt{2}, 30+20\sqrt{2} \right ) \approx \left ( -1.6, 58.3\right )$

Using the distance formula for this point from the origin gives us a distance of ~58 miles.

Now for the time. We traveled 30 miles at 40 mph. This means we traveled for $\frac{30}{40}$ hours or 45 minutes.

We then traveled 40 mph for 40 minutes. Increasing our total time traveled to 85 minutes.

Finally, we traveled 40 miles in 25 minutes, leaving our total time traveled at 110 minutes. Returning to hours, we have $\frac{110}{60}$ hours or 1.83 hours.

Our final average speed traveled is then:

$\frac{58 miles}{1.83 hours} \approx 31.7 mph$

The closest answer to this value is 32 mph.

What is the distance between $(3,4)$ and $(8,16)$ ?

$13$

$12$

$20$

$23$

$17$

Explanation

The formula for the distance between two points is $d = \sqrt{(x_{2} - x_{1})^2+(y_{2} - y_{1})^2}$ .

Plug in the points:

$d=\sqrt{(8-3)^2+(16-4)^2}$

$d = \sqrt{5^2+12^2} = 13$

Give the length, in terms of , of a segment on the coordinate plane whose endpoints are and $\left ( \frac{1}{2}, \frac{1}{2} \right )$ .

$\sqrt{ 2 N^{2} - 2 N + \frac{1}{2} }$

$\sqrt{ 2 N^{2} + 2 N + \frac{1}{2} }$

$\sqrt{ 2 N^{2} - \frac{1}{2} }$

$\sqrt{ 2 N^{2} + \frac{1}{2} }$

$\sqrt{ 2 N^{2} + 2 N }$

Explanation

The length of a segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be calculated using the distance formula:

$d = \sqrt{(x_{2}- x_{1})^{2} + (y_{2}- y_{1})^{2} }$

Setting $x_{2} = y_{2} = N$ and $x_{1} = y_{1} = \frac{1}{2}$ and substituting:

$d = \sqrt{ \left (N- \frac{1}{2}\right ) ^{2} + \left (N- \frac{1}{2}\right ) ^{2}}$

The binomials can be rewritten using the perfect square trinomial pattern:

$d = \sqrt{ \left [N^{2} - 2 \cdot N \cdot \frac{1}{2} + \left ( \frac{1}{2}\right ) ^{2} \right ] + \left [N^{2} - 2 \cdot N \cdot \frac{1}{2} + \left ( \frac{1}{2}\right ) ^{2} \right ] }$