SSAT Upper Level Quantitative - How to find the length of a line with distance formula

One leg of a triangle has endpoints at the coordinates $(-1, -2)\text{ and }(-4, -1)$ . Find the length of this leg.

$\sqrt{10}$

$2\sqrt2$

Explanation

Use the distance formula to find the length of the leg.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Where,

and .

$\text{Length}=\sqrt{(-4-(-1))^2+(-1-(-2))^2}=\sqrt{9+1}=\sqrt{10}$

A segment on the coordinate plane whose endpoints are and the origin has length 10. Calculate .

$A = \pm 5 \sqrt{2}$

$A = \pm 5$

$A = \pm 10$

$A = \pm 5 \sqrt{3}$

$A = \pm 10 \sqrt{2}$

Explanation

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

If we set $x_{1}= y_{1} = 0, x_{2}= y_{2} = A, d=10$ , we can solve for :

$10= \sqrt{\left (A-0 \right )^{2}+\left (A-0 \right )^{2}}$

$10= \sqrt{ A ^{2}+A ^{2}}$

$10= \sqrt{ 2A ^{2}}$

$10^{2}=\left ( \sqrt{ 2A ^{2}} \right )^{2}$

$100 = 2A^{2 }$

$100 \div 2 = 2A^{2 } \div 2$

$50 = A^{2}$

$A = \pm \sqrt{50 } = \pm \sqrt{25} \cdot \sqrt{2} = \pm 5 \sqrt{2}$

Give the length of a segment on the coordinate plane whose endpoints are $\left (\frac{1}{5} , \frac{4}{5} \right )$ and $\left ( -\frac{3}{5}, \frac{1}{5} \right )$ .

The correct answer is not among the other responses.

$\frac{1}{5}$

$\frac{\sqrt{1}}{5}$

$\frac{\sqrt{18}}{5}$

$\frac{\sqrt{32}}{5}$

Explanation

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

Set $x_{1}= \frac{1}{5}, y_{1}= \frac{4}{5}, x_{2}= -\frac{3}{5}, y_{2} = \frac{1}{5}$ :

$d = \sqrt{\left ( \frac{1}{5}- \left ( - \frac{3}{5}\right ) \right )^{2}+\left (\frac{4}{5}-\frac{1}{5} \right )^{2}}$

$= \sqrt{\left ( \frac{4}{5} \right )^{2}+\left (\frac{3}{5} \right )^{2}}$

$= \sqrt{ \frac{16}{25} + \frac{9}{25} }$

$= \sqrt{ \frac{25}{25} }$

$= \sqrt{1}$

This is not given among the responses.

A line segment has endpoints at $(-2, -1)\text{ and }(8, 1)$ . Find the length of this line.

$2\sqrt{26}$

$6\sqrt{3}$

Explanation

Use the distance formula to find the length of the line segment.

$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Where,

and .

$\text{Distance}=\sqrt{(8-(-2))^2+(1-(-1))^2}=\sqrt{100+4}=\sqrt{104}=2\sqrt{26}$

A segment on the coordinate plane whose endpoints are and the origin has length 10. If is assumed to be positive, calculate .

$A = 2\sqrt{5}$

$A = 4\sqrt{5}$

The correct answer is not given among the other responses.

Explanation

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

If we set $x_{1}= y_{1} = 0, x_{2}=2A, y_{2} = A, d=10$ , we can solve for :

$10= \sqrt{\left (2A-0 \right )^{2}+\left (A-0 \right )^{2}}$

$10= \sqrt{\left (2A \right )^{2}+A^{2}}$

$10= \sqrt{4A ^{2}+A^{2}}$

$10= \sqrt{5A ^{2} }$

$10^{2}=\left ( \sqrt{ 5A ^{2}} \right )^{2}$

$100 = 5A^{2}$

$100\div 5 = 5A^{2} \div 5$

$20 = A^{2}$

$A = \sqrt{20} = \sqrt{4} \cdot \sqrt{5} = 2\sqrt{5}$

If the two points on a line are $“\left(“\frac{1}{3}, \frac{1}{3}“\right)”$ and , what is the approximate distance connecting the two points?

Explanation

Write the distance formula.

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Substitute the point values.

$d=\sqrt{\left( 4-\frac{1}{3}\right)^2+\left( 2-\frac{1}{3}\right)^2} = \sqrt{\left( \frac{11}{3}\right)^2+\left( \frac{5}{3}\right)^2}$

$d=\sqrt{13.\bar{4}+2.\bar{7}}= \sqrt{16.\bar{2}}= 4.027$

The approximated distance is 4.

Give the length, in terms of , of a segment on the coordinate plane whose endpoints are and

$\sqrt{2A^{2}-12A+18 }$

$\sqrt{2A^{2} +18 }$

$\sqrt{2A^{2} -18 }$

$\sqrt{2A^{2}+12A+18 }$

Explanation

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

Setting $x_{1}= y_{1} =3, x_{2}= y_{2} = A$ , we can find our expression:

$d = \sqrt{\left (A-3 \right )^{2}+\left (A-3\right )^{2}}$

$= \sqrt{\left (A^{2}-6A+9 \right ) + \left (A^{2}-6A+9 \right ) }$

$= \sqrt{2A^{2}-12A+18 }$

Give the length, in terms of , of a segment on the coordinate plane whose endpoints are and the origin.

$\sqrt{ 2A^{2}+ 8 }$

$\sqrt{ 2A^{2}+ 8A+8 }$

$\sqrt{ 2A^{2}- 8A+8 }$

$2 \sqrt{2A}$

$\sqrt{ 2A^{2}- 8 }$

Explanation

The distance between two points $(x_{1}, x_{2})$ and $(y_{1}, y_{2})$ can be obtained using the formula

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

Setting $x_{1}= y_{1} = 0, x_{2}=A+2, y_{2} = A-2$ , we can find our expression:

$d = \sqrt{\left ( \left [A+2 \right )-0\right ]^{2}+\left [\left (A-2 \right )-0 \right ] ^{2}}$ $= \sqrt{\left ( A+2 \right ) ^{2}+ \left (A-2 \right ) ^{2}}$

$= \sqrt{\left ( A^{2}+4A+4 \right )+\left ( A^{2}-4A+4 \right ) }$

$= \sqrt{ 2A^{2}+ 8 }$

A side of a square is graphed onto a coordinate plane. The side has endpoints at $(-3, 1)\text{ and }(4, -1)$ . Find the length of the side of the square.