Question

please i need help with this

Answer 1

Given :-

• Line p has an equation of y = 5/3x - 4 .
• Line q includes point (-10,-3) and is perpendicular to the line p .

To Find :-

• The equation of line q .

Solution :-

The equation of the line p is ,

$\sf\blue{\longrightarrow}$ y = 5/3x - 4

On comparing to slope intercept form of the line which is y = mx + c , we have ,

$\sf\blue{\longrightarrow}$ m = 5/3

Now as we know that the product of slopes of two perpendicular lines is -1 . So the slope of the perpendicular line will be ,

$\sf\blue{\longrightarrow} m_{\perp} = -3/5$

Now here the line q passes through the point (-10,-3) . So on using the point slope form of the line we get ,

$\sf\blue{\longrightarrow} y - y_1 = m(x-x_1)$

$\sf\blue{\longrightarrow}$ y - (-3) = -3/5[ x-(-10)]

$\sf\blue{\longrightarrow}$ y +3 = -3/5(x+10)

$\sf\blue{\longrightarrow}$ y+3 = -3/5x - 6

$\sf\blue{\longrightarrow}$ y +9 = -3/5x

$\sf\blue{\longrightarrow}$ y = - 3/5x - 9

Hence the required answer is y = -3/5x - 9 .

Answer 2

Answer:

For Perpendicular lines you have to know

Step-by-step explanation:

Slope of P ×slope of q=-1 (always since they are perpned..)

so as P has the slope 5/3 in order to optain - 1 the slope of q should be negative of the inverse of p

so q solpe =-3/5 then lets try (5/3)×(-3/5)? =-1 yes

And be aware that you have to know the formula of straight line : y-ypoint=slope(x-xpoint) or y=ax+b

so y+3=-3/5(x+10) ===> y=-3/5x - 30/5 - 3

uniting the denominater ==>y=-3/5x - 45/5

so that q equation is :y=-3/5x - 9