Question

which linear equation has the points $\left(\frac{1}{2}, \frac{3}{4}\
ight)$ and $(4, 3)$ as solutions? *
$9x - 14y = -6$
$9x + 14y = 78$
$y - 3 = \frac{14}{9}(x - 4)$
$y - 4 = \frac{9}{14}(x - 3)$

Explanation:

Step1: Calculate slope of the line

The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$. Let $(x_1,y_1)=(\frac{1}{2},\frac{3}{4})$ and $(x_2,y_2)=(4,3)$.

$$ m = \frac{3 - \frac{3}{4}}{4 - \frac{1}{2}} = \frac{\frac{9}{4}}{\frac{7}{2}} = \frac{9}{4} \times \frac{2}{7} = \frac{9}{14} $$

Step2: Use point-slope form

Point-slope form is $y - y_2 = m(x - x_2)$, using $(4,3)$:

$$ y - 3 = \frac{9}{14}(x - 4) $$

Step3: Verify with standard form

Rearrange to standard form $Ax+By=C$:

$$ y - 3 = \frac{9}{14}x - \frac{36}{14} \\ 14y - 42 = 9x - 36 \\ 9x - 14y = -6 $$

Step4: Check the options

Match the derived equation to the given choices.

Answer:

9x - 14y = -6