Question

Find an equation for the line with the given property

(a) It passes through the point $\left(3,-6\right)$and is parallel to the line $3x+y-10=0$

(b) It has x-intercept 6 and y-intercept

Short answer

(a) Equation of the line is

$3x+y-3=0$

(b) Equation of the line is

$2x+3y-12=0$

Step-by-step solution

Part a. Step 1. Definition of parallel lines.

Two non- vertical lines are parallel if and only if they have the same slope

Part a. Step 2. Determining the slope.

The given equation is

$3x+y-10=0$

$y=-3x+10$subtract 3x

So, the line has slopewidth="56">m=-3

Part a. Step 3. Determining the equation of the line.

Since the line is parallel to given equation whose slope is $-3$

So, the new equation slope is $-3$and it passes through $(3,-6)$

$$\begin{gathered} y+6=-3(x-3) \\ \Rightarrow y+6=-3x+9 \\ \Rightarrow y+3x-3=0 \end{gathered}$$

Part b. Step 1. Definition of x-intercepts.

The x-coordinates of points where the graph of an equation intersects the x-axis.

Part b. Step 2. Definition of y-intercepts.

The y-coordinates of points where the graph of an equation intersects the y-axis.

Part b. Step 3. Determining the equation of the line.

So, $\frac{x}{a}+\frac{y}{b}=1$

Where a is the x-intercept and b is the y- intercept

$\frac{x}{6}+\frac{y}{4}=1$

$\Rightarrow \frac{2x+3y}{12}=1$

$\therefore 2x+3y-12=0$