Question

HELPPPPPPP PLEASEEE

Answer 1

Step-by-step explanation:

Hey there!

Given,

The line is passing through point (-1,-2) and perpendicular to the line (y+6) = 4/5(x-5).

Now,

The equation of a st.line passing through point (-1,-2) is,

$(y - y1) = m1(x - x1)$

Put all values.

$(y + 2) = m1(x + 1)$

It is the 1st equation.

Another equation is (y+6)= 4/5(x-5).

$y = \frac{4}{5} x - 10$

From equation (ii) {After comparing the equation with y =mx+c}.

We get,

M2= 4/5.

Now,

As per the condition of perpendicular lines,

m1×m2= -1.

$m1 \times \frac{4}{5} = - 1$

Simplify them to get answer.

$4m1 = - 5$

Therefore the slope is, -5/4.

Now, keep the slope in 1st equation.

$(y + 2) = \frac{ - 5}{4} (x + 1)$

Simplify them to get answer.

$(y + 2) = \frac{ - 5}{4} x - \frac{5}{4}$

$y = \frac{ - 5}{4} x - \frac{5}{4} - 2$

$y = \frac{ - 5}{4} x - \frac{13}{4}$

Therefore the required equation is y = -5/4x -13/2.

Hopeit helps....

Answer 2

Answer:

y = - $\frac{5}{4}$ x - $\frac{13}{4}$

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

y + 6 = $\frac{4}{5}$(x - 5) ← is in point- slope form

with slope m = $\frac{4}{5}$

Given a line with slope m then the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{\frac{4}{5} }$ = - $\frac{5}{4}$

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = - $\frac{5}{4}$ , thus

y = - $\frac{5}{4}$ x + c ← is the partial equation

To find c substitute (- 1, - 2) into the partial equation

- 2 = $\frac{5}{4}$ + c ⇒ c = - 2 - $\frac{5}{4}$ = - $\frac{13}{4}$

y = - $\frac{5}{4}$ x - $\frac{13}{4}$ ← equation of perpendicular line