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write an equation of the line through (-3, -6) having slope \\(\\frac{1…

Question

write an equation of the line through (-3, -6) having slope \\(\frac{13}{17}\\). give the answer in standard form.

the equation of the line is \\(\square x - \square y = \square\\).

Explanation:

Step1: Use point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope. Here, $x_1=-3$, $y_1 = - 6$ and $m=\frac{13}{17}$.
Substitute these values into the point - slope form:
$y-(-6)=\frac{13}{17}(x - (-3))$
Simplify the left - hand side and the right - hand side:
$y + 6=\frac{13}{17}(x + 3)$

Step2: Eliminate the fraction

Multiply both sides of the equation by 17 to get rid of the fraction:
$17(y + 6)=13(x + 3)$

Step3: Expand both sides

Expand the left - hand side: $17y+102$
Expand the right - hand side: $13x + 39$
So, $17y+102 = 13x+39$

Step4: Rearrange into standard form $Ax - By = C$

Subtract $17y$ from both sides: $102=13x-17y + 39$
Subtract 39 from both sides: $13x-17y=102 - 39$
Calculate $102-39 = 63$. So, $13x-17y=63$

Answer:

The equation of the line is $\boldsymbol{13}x-\boldsymbol{17}y=\boldsymbol{63}$.