SAT Lines, Slopes and Intersections

Correct Answer: E

Solution 1:

Tip: The slope of a line is defined as $\frac{\text{change in}\ y}{\text{change in}\ x}.$
Option (E) says that the slope of the line that passes through points $(x,y)$ and $(1,3)$ is equal to the slope of the line that passes through the points $(x,y)$ and $(-2, 5)$ . Thus, this point must lie on the line between $(1,3)$ and $(-2, 5)$ , which gives us the answer.

Solution 2:

Tip: The slope of a line is defined as $\frac{\text{change in}\ y}{\text{change in}\ x}.$
Tip: Point-slope form: $y-y_{1}=m(x-x_{1}),$ where $m$ is the line's slope, and $(x_{1}, y_{1})$ is a point on the line.
Let's find the equation of the line. The slope of the line is equal to

$m = \frac{\text{change in}\ y}{\text{change in}\ x} = \frac{ 5 - 3 } { -2 - 1 } = - \frac{ 2 } { 3} .$

By the point-slope form, the equation of this line is

$\begin{array} { r c l l } ( y-3) &=& - \frac{2}{3} ( x -1 ) &\quad \text{point-slope form} \\ 3 ( y-3) &=& -2( x -1 ) &\quad \text{multiply both sides by}\ 3 \\ 3y - 9 &=& - 2x + 2 &\quad \text{expand terms} \\ 3y + 2x - 11 &=& 0 &\quad \text{simplify} \\ \end{array}$

Now, we check this equation against the answer choices.

(A) $y = -2x + 5$ is not the same as $3y + 2x - 11 = 0$ .

(B) $y = 2x + 1$ is not the same as $3y + 2x - 11 = 0$ .

(C) If $\frac{ y-3} { x-1} = \frac{2}{3}$ , then cross multiplying gives us $3 ( y-3) = 2 ( x-1)$ . Expanding gives $3y - 9 = 2x - 2$ . Simplifying gives $3y - 2x - 7 = 0$ . This is not the same as $3y + 2x - 11 = 0$ .

(D) If $\frac{ y - 5 } { x + 2 } = - \frac{3}{2}$ , then cross multiplying gives $2 ( y- 5) = -3 ( x + 2)$ . Expanding gives $2y - 10 = -3x - 6$ . Simplifying gives $2y + 3x - 4 = 0$ . This is not the same as $3y + 2x - 11 = 0$ .

(E) If $\frac{ y-3}{x-1} = \frac{ y-5}{x+2}$ , then cross multiplying gives $(x+2)(y-3) = (x-1)(y-5)$ Expanding gives $xy + 2y - 3x - 6 = xy - y - 5x + 5$ . Simplifying gives $3y + 2x - 11 = 0$ . Hence, this is the answer.

Solution 3:

Tip: Plug and check.
Each of the two points, when plugged into the equation of the line that passes through them should yield a true statement. So we plug and check. Here, for each wrong answer, we only dothe work for the point that yields a contradiction.

(A) $-2 \times (-2) + 5 = 9 \neq 5$ so $(-2, 5 )$ does not lie on the line.

(B) $2 \times (-2) + 5 = 1 \neq 5$ so $(-2, 5 )$ does not lie on the line.

(C) $\frac{ 5 - 3 } { -2 - 1} = - \frac{2}{3} \neq \frac{2}{3}$ , so $(-2, 5)$ does not lie on the line.

(D) $\frac{ 3-5 } {1 + 2 } = \frac{ -2}{ 3 } \neq - \frac{3}{2}$ , so $(1,3)$ does not lie on the line.

We are left with option (E), which has to be the answer.

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Incorrect Choices:

(A) , (B) , (C) and (D)
Look at Solution 3 for how to eliminate these choices by plugging and checking.