Question

Write an equation for the line in point/slope form and slope/intercept form that has the given condition.

6. Passes through (3,2) and is parallel to 2x-y=4

7. Passes through (-1,-1) and is perpendicular to y=5/2x+3

Answer 1

Write an equation for the line in point/slope form and slope/intercept form that has the given condition.

6. Passes through (3,2) and is parallel to $2x - y = 4$

7. Passes through (-1,-1) and is perpendicular to $y = \frac{5}{2} x + 3$

**Solution:**

6. The slope of the line is $m = 2$

In the point/slope form we have $y - y_{1} = m(x - x_{1})$

$$y - 2 = 2 (x - 3)$$

In the slope/intercept form $y = 2x + b$

If it passes through (3,2) then $2 = 2 * 3 + b \quad \Rightarrow \quad b = -4$

So in the slope/intercept form we have: $y = 2x - 4$

Answer: $y - 2 = 2(x - 3)$

$$y = 2 x - 4$$

7. The slope of the line is $m = -\frac{2}{5}$ (from the condition of perpendicularity)

In the point/slope form we have $y - y_{1} = m(x - x_{1})$

$$y + 1 = - \frac {2}{5} (x + 1)$$

In the slope/intercept form $y = 2x + b$

If it passes through $(-1, -1)$ then $-1 = -\frac{2}{5} * (-1) + b \quad \Rightarrow \quad b = -\frac{7}{5} = -1\frac{2}{5}$

So in the slope/intercept form we have: $y = -\frac{2}{5} x - 1\frac{2}{5}$

Answer: $y + 1 = -\frac{2}{5} (x + 1)$

$$y = - \frac {2}{5} x - 1 \frac {2}{5}$$