Which of the following is an equation of the line in the $x y$ plane with $x$ -intercept $\frac{5}{3}$ and $y$ intercept $\frac{10}{7}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-7 y+6 x=10$ $\newline$ (B) $7 y+6 x=-10$ $\newline$ (C) $7 y+6 x=10$ $\newline$ (D) $7 y-6 x=10$

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Algebra 1

Write a quadratic function from its x-intercepts and another point

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Q. Which of the following is an equation of the line in the

x y

plane with

x

-intercept

\frac{5}{3}

and

y

intercept

\frac{10}{7}

\newline

Choose

1

answer:

\newline

(A)

-7 y+6 x=10

\newline

(B)

7 y+6 x=-10

\newline

(C)

7 y+6 x=10

\newline

(D)

7 y-6 x=10

Finding Intercepts: The $x$ -intercept of a line is the point where the line crosses the $x$ -axis, which means the $y$ -coordinate is $0$ . Similarly, the $y$ -intercept is where the line crosses the $y$ -axis, meaning the $x$ -coordinate is $0$ . We can use these intercepts to find the equation of the line in the form of $Ax + By = C$ , where $A$ , $x$ $0$ , and $x$ $1$ are constants.
Substituting x-intercept: Given the x-intercept $(\frac{5}{3}, 0)$ , we can substitute $x = \frac{5}{3}$ and $y = 0$ into the equation $Ax + By = C$ to find the relationship between $A$ and $C$ .
$A(\frac{5}{3}) + B(0) = C$
$\frac{5A}{3} = C$
Substituting y-intercept: Given the y-intercept $(0, \frac{10}{7})$ , we can substitute $x = 0$ and $y = \frac{10}{7}$ into the equation $Ax + By = C$ to find the relationship between $B$ and $C$ .
$A(0) + B\left(\frac{10}{7}\right) = C$
$\frac{10B}{7} = C$
Equating the expressions for C: We now have two equations relating $A$ , $B$ , and $C$ : $\newline$ $\frac{5A}{3} = C$ $\newline$ $\frac{10B}{7} = C$ $\newline$ We can equate these two expressions for C to find the relationship between $A$ and $B$ . $\newline$ $\frac{5A}{3} = \frac{10B}{7}$ $\newline$ Multiplying both sides by $21$ (the least common multiple of $3$ and $7$ ) to clear the fractions, we get: $\newline$ $35A = 30B$
Solving for A and B: To find $A$ and $B$ , we can choose a common multiple of $35$ and $30$ that will make $A$ and $B$ integers. The smallest such number is $210$ , so we can set $35A = 210$ and $30B = 210$ . Solving for $A$ and $B$ gives us: $\newline$ $B$ $1$ $\newline$ $B$ $2$
Solving for C: Now that we have $A$ and $B$ , we can use either of the two equations we found earlier to solve for $C$ :
$\frac{5A}{3} = C$
$\frac{5(6)}{3} = C$
$C = 10$
Final Equation: We have found $A = 6$ , $B = 7$ , and $C = 10$ . The equation of the line is therefore: $\newline$ $6x + 7y = 10$
Comparing with answer choices: We can now compare the equation we found with the answer choices given: $\newline$ (A) $-7y + 6x = 10$ $\newline$ (B) $7y + 6x = -10$ $\newline$ (C) $7y + 6x = 10$ $\newline$ (D) $7y - 6x = 10$ $\newline$ The correct equation that matches our derived equation is (C) $7y + 6x = 10$ .