Given:
Equation of line 
.
To find:
The equation of line that goes through the point ( − 21 , 2 ) and is perpendicular to the given line.
Solution:
The given equation of line can be written as
Slope of line is
Product of slopes of two perpendicular lines is -1. So, slope of perpendicular line is
![[\because m_1=\dfrac{7}{4}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20m_1%3D%5Cdfrac%7B7%7D%7B4%7D%5D)
Now, the slope of perpendicular line is 
and it goes through (-21,2). So, the equation of line is
Therefore, the required equation in slope intercept form is .
Answer:
3) 9.2ml
4) .25ml
Have to have 20 characters
It should be x < -10 because when you divide by a negative number the inequality sign flips.
Answer:
P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.
Step-by-step explanation:
P and Q are the intersection points of
x-y+1 = 0 and the circle x^2 + y^2 = 25
sub y = x+1 into the circle
x^2 + (x+1)^2 = 25
x^2 + x^2 + 2x + 1 - 25 = 0
x^2 + x - 12 = 0
(x+4)(x-3) = 0
x = 3 or x = -4
y = 4 or y = -3
so P(3,4) and Q(-4,3) are our two points
Height of triangle.
h = |0 - 0 + 1|/√2 = 1/√2
PQ = √( (-7)^2 + 1^2) = √50 = 5√2
area POQ = (1/2)(1/√2)(5√2) = 5/2 square units
hope this helped
Answer:
freeeeeeeee ahhh
Step-by-step explanation:
:)
